Answer:
10,200 N
Explanation:
Given:
Mass of the car m = 1200 kg
Acceleration of the car a = 8.5 m/s^2
Force F=?
[tex]\because F = ma\\
\therefore F = 1200 \times 8.5\\
\therefore F = 10,200 N \\ [/tex]
Your shopping cart has a mass of 65 kilograms.In order to accelerate the shopping cart down an aisle at 0.30 m/s^2, what Force would you need to use or apply to the cart assuming the coefficient of friction between the cart and the floor is 0.01?
Answer:
25.88 N
Explanation:
Mass of the shopping cart,[tex]m=65 kg[/tex]
The coefficient of friction between the cart and the floor, [tex]\mu= 0.01[/tex]
Let, F be the required force to accelerate the cart at [tex]a=0.30 m/s^2[/tex].
Gravitational force, mg, acts downward which is being balanced by the normal reaction, N, on the cart by the floor.
As the motion of the cart in the vertical direction is zero. So, using the static equilibrium condition will be zero.
From free body diagram (FBD):
[tex]N-mg=0[/tex]
[tex]\Rightarrow N=mg.[/tex]
the net force action in this direction The frictional force, f, acts in the direction to opposes the motion of the cart as shown.
[tex]f=\muN=\mu mg[/tex]
Now, apply the dynamic equilibrium condition in the horizontal direction, i.e. net force acting on the body equals the rate of change of momentum of the body. From the FBD of the cart, we have
[tex]F-f-ma=0[/tex]
[tex]\Rightarrow F=f+ma[/tex]
[tex]\Rightarrow F=\mu mg+ma[/tex]
[tex]\Rightarrow F=m(\mu g +a)[/tex]
[tex]\Rightarrow F=65(0.01\times 9.81 + 0.3)[/tex]
[tex]\Rightarrow F=25.88 N.[/tex]
Hence, 25.88 N force required to accelerate the body with 0.03 [tex]m/s^2[/tex] .
Amass oscillates in simple harmonic motion with amplitude A , if the mas is halved , but the amplitude is not changed , what will happen to the total mechanical energy of the system
A- It will increase
B-it will decrease
c-It will stay at the same
D-None of these
NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
Answer:
Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.
We are given:
initial velocity (u) = 0 m/s [vertical]
final velocity (v) = v m/s [vertical]
time taken to reach the ground (t) = t seconds
acceleration (a) = 10 m/s/s [vertical , due to gravity]
height from the ground (h) = 130 m
displacement (s) = 53 m [horizontal]
Solving for time taken:
From the third equation of motion:
s = ut + 1/2 at²
130 = (0)(t) + 1/2 * (10) * t²
130 = 5t²
t² = 26
t = √26 seconds or 5.1 seconds
Final Horizontal velocity of the ball
Since the horizontal velocity of the ball will remain constant:
the ball covered 53 m in 5.1 seconds [horizontally]
horizontal velocity of the ball = horizontal distance covered / time taken
Velocity of the ball = 53 / 5.1
Velocity of the ball = 10.4 m/s
Answer:
51.51519 m/s
Explanation:
Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]
X-direction | Y-direction
[tex]x=x_{o} +v_{xo}t[/tex] | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]
[tex]53=0v_{xo}(5.15078)[/tex] | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]
[tex]53=v_{xo} (5.15078)[/tex] | [tex]-130=-4.9t^2[/tex]
[tex]\frac{53}{5.15078} =v_{xo}[/tex] | [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]
[tex]10.2897=v_{xo}[/tex] | [tex]5.15078=t[/tex]
[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex] | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]
[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]
[tex]v=51.51519 m/s[/tex] | [tex]v_{y}=50.47771[/tex]
6. What velocity vector will move you 200 miles east in 4 hours travelingat a constant speed?
Answer:
50 mph east
Explanation:
A car traveling 21 m/s is accelerated uniformly at the rate of 2.2 m/s ^2 for 6.9 s. What is the car’s final speed?
Answer:
36m/s
Explanation:
v=u+at
v=21+(2.2×6.9)
v=21+15.18
v=36.18
v=36m/s
Reset Answers
Practice Energy transformations (Ouestion 5 of 10)
Q5
A penny is dropped from the top of the Statue of Liberty. The Statue of Liberty is 93 m tall If a penny has about 3 Joules of gravitational potential energy at the top, how much
potential energy will have transformed into kinetic energy at the half way point (46.5m) ?
Explanation:
At a height of 93 m, the gravitational potential energy is given by :
P = mgh
Where, m is the mass of a penny
We can find its mass.
[tex]m=\dfrac{P}{gh}\\\\m=\dfrac{3}{9.8\times 93}\\\\m=0.00329\ kg[/tex]
We need to find how much potential energy will have transformed into kinetic energy at the half way point (46.5m). It can be calculated as :
[tex]E=mgh'\\\\E=0.00329\times 9.8\times 46.5\\\\E=1.499\ J[/tex]
Hence, a penny will have transferred 1.499 J of potential energy into kinetic energy.
NO ONE WILL HELP, PLEASE PLEASE HELP, I HAVE AN HOUR TO GET 4 PAGES DONE! Find the average speed of a marble that takes 6 seconds to roll 30 m across a gymnasium floor.
An African Swallow can travel at an average velocity of 11 m/s. How far can an African Swallow carry a coconut in 120 seconds?
Answer:
The distance will be x = 1320 [m]
Explanation:
To solve this problem we must use the expression of physics that relates space to time, which is defined as speed.
v = Speed = 11 [m/s]
t = time = 120 [s]
x = distance [m]
v = x/t
x = v*t
x = 11*120
x = 1320 [m]
A car travels 45 km due north and 70 km west. What is the car's displacement?
6 points
24.7 km northeast
83.2 km northwest
76.5 km northwest
115 km north
Answer:
A
Explanation:
I need help with this answer
Answer:
Single Replacement
Explanation:
Anuja hit a golf ball on a level field at 70 degrees and 40 degrees with the same total speed as shown below.
70°
40°
Which launch angle causes the ball to be in the air for the longest time?
o not enough information
40 degrees
70 degress
times are the same
Answer:
At 40 deg Vy = V sin 40
at 70 deg Vy = V sin 70
So the ball launched at 70 deg has the greatest vertical velocity and will remain in the air the longest:
Since t = Vy / g time for to reach zero vertical velocity and also the time for the ball to reach velocity Vy on the downward path
Mathis kicked a ball on a level surface at 30∘ and 60∘ with the same total speed as shown below.
Which launch angle results in the greater maximum height for the ball?
Answer: CORRECT (SELECTED)
60
A +5.00 pC charge is located on a sheet of paper.
(a) Draw to scale the curves where the equipotential surfaces due to these charges intersect the paper. Show only the surfaces that have a potential (relative to infinity) of 1.00 V, 2.00 V, 3.00 V, 4.00 V, and 5.00 V.
(b) The surfaces are separated equally in potential. Are they also separated equally in distance?
(c) In words, describe the shape and orientation of the surfaces you just found.
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = [tex]q_{int}[/tex] /ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
2. What is the normal force acting a 800-kg car if there are two 55-kg people sitting inside the car?
What is the speed of a car that travels 0.200 km in 95 seconds? State your answer in meters per second.
Answer:
2.10 meters per secondExplanation:
speed (velocity) = distance / time
= 0.2 km (1000 m/km)
95 secs.
= 2.10 meters per second
A boy kicked off a cliff and lands 151m away 45s later. What was the initial velocity? How tall is the cliff?
This chart shows Dan's budget:
Did Dan stay on budget? Why or why not?
Amount budgeted
tem
nome
Food
Rent
Debonary spending
Income
$100
S500
$100
5750
Amount spent
55
S90
3500
5140
Yes, Dan spent as much as he earned.
No, Dan should move to a new apartment
O Yes, Dan uses his savings to cover extra expense
No, Dan should reduce his discretionary spending,
Answer
No dan should reduce his discretionary apendings.
Explanation:
classify the planets as inner planets of outer planets explain your answer
Answer: Planets A and B have a rocky mantle and an iron core. They are at a distance of 1 AU or less from the Sun. This means they are relatively close to the Sun. There are the properties of inner (terrestrial) planets. So planet A and B are inner planets.
Explanation:
Planet C is composed if the gases hydrogen and helium. It has a high mass and is much farther from 1 AU from the sun. These properties are consistent with the outer planets. So, planet C is an outer (gas giant) planet.
Answer:
Planets A and B have a rocky mantle and an iron core. They are at a distance of 1 AU or less from the Sun. This means they are relatively close to the Sun. There are the properties of inner (terrestrial) planets. So planet A and B are inner planets.
Explanation:
• An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is expected that:
(a) the coin will reach the bottom first.
(b) the feather will reach the bottom first.
(c) both objects will reach the bottom at the same instant.
• If this experiment is repeated at a place 2000 kilometers above the sea-level, the acceleration due to gravity gexp is expected to :
(a) increase. (b) decrease. (c) remain constant.
Answer:
c-) both objects will reach the bottom at the same instant.
(b) decrease.
Explanation:
Although the feather is lighter than the coin, the tube where the experiment is performed is evacuated. Therefore there is no air that prevents the feather from falling freely with the same acceleration and speed as the coin.
In fact in the equations of kinematics proposed by Newton, the mass of the bodies is not taken into account, as we can see in the following equation:
[tex]v_{f}= v_{i} +g*t[/tex]
where:
Vf = final velocity [m/s]
Vi = initial velocity [m/s]
g = gravity acceleration [m/s^2]
t = time [s]
Therefore the answer is C.
Gravitational pull is a function of height, as the height of the body increases, the force of gravity decreases.
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?
Answer:
1 = 5.4 J
2 = 0.1979 C
3 = 5
Explanation:
Energy in a capacitor, E is
E = 1/2 * C * V²
E = 1/2 * 3000*10^-6 * 60²
E = 1/2 * 3000*10^-6 * 3600
E = 1/2 * 10.8
E = 5.4 J
E = Q²/2C = 6.53 J
E * 2C = Q²
Q² = 6.53 * 2 * 3000*10^-6
Q² = 13.06 * 3000*10^-6
Q² = 0.03918
Q = √0.03918
Q = 0.1979 C
The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.
12) If a man weighs 900 N-on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s?
Answer:
Force=Mass*acceleration
on earth, acceleration=9.81 m/s^2
900 N=Mass*9.81 m/s^2
Mass=91.74 Kg
F=Mass*acceleration(Jupiter)
F=91.74Kg*25.9m/s
F=2376.066 N on Jupiter
Plz mark me as brainliest if u found it helpful
The displacement of an object moving 330 km North for 2 hours and an additional 220 km North for 5 hours is?
Answer:
Usbe
Explanation:
A sound wave in a steel rail has a frequency of 620 hz and a wavelength of 10.5m. What is the speed of sound in steel?
What is the average speed of the whole trip?
Answer:
65 miles per hour.
Explanation:
Answer:
Average speed of entire trip is : 3.67 m/s
Explanation:
Speed is calculated as the distance covered divided by time
Average speed will be the :speed before stop add to speed after stopping then divide by 2
Speed before stop = 200/ 60 = 3.33 m/s
Speed after stop = 400/100= 4m/s
Average speed = (3.33+4)/2 = 7.33/2 = 3.67 m/s
A 2.60-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.380. Determine the kinetic frictional force that acts on the box for each of the following cases. (a) The elevator is stationary. N (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2. N (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.20 m/s2. N Additional Materials
Answer:
a) F = 9.69 N
b) F = 10.88 N
c) F = 8.51 N
Explanation:
a) The kinetic frictional force when the elevator is stationary is the following:
[tex] F_{k} = \mu_{k}N = \mu(mg) [/tex]
Where:
F(k) is the kinetic frictional force
N is the normal force = mg
m: is the mass = 2.60 kg
g: is the gravity = 9.81 m/s²
μ(k) is the coefficient of kinetic friction = 0.380
[tex] F_{k} = \mu(mg) = 0.380*2.60 kg*9.81 m/s^{2} = 9.69 N [/tex]
b) When the elevator is accelerating upward with acceleration "a" equal to 1.20 m/s²
[tex] F_{k} = \mu_{k}N = \mu[m(g + a)] [/tex]
The normal is equal to mg plus ma because the elevator is accelerating upward
[tex] F_{k} = \mu m(g + a) = 0.380*2.60 kg(9.81 m/s^{2} + 1.20 m/s^{2}) = 10.88 N [/tex]
c) When the elevator is accelerating downward with a =1.20 m/s² we can find the kinetic frictional force similar to the previous case:
[tex] F_{k} = \mu m(g - a) = 0.380*2.60 kg(9.81 m/s^{2} - 1.20 m/s^{2}) = 8.51 N [/tex]
I hope it helps you!
David drove the first 6 hours of his journey at 65km/hr and the last 3 hours of his journey at 80km/hour. How far is the whole journey in km?
630 kilometers for the whole journey
a satellite is revolving around the sun in a circular orbit with uniform velocity v. if the gravitational force suddenly disappears the velocity of the satellite will be?
Answer:
when gravitational force suddenly disappears, then only centrifugal force will be acting and velocity is tangential to the orbit and hence, the satellite will fly off tangentially with same velocity v.
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1
Complete Question
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
[tex]A.\ \ P_2=4P_1[/tex]
[tex]B.\ \ P_2=\frac{4}{3} P1[/tex]
[tex]C.\ \ P_2=P_1[/tex]
[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]
[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]
Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the brick is M
The height height of the 10th floor is H = 4y
The height attained during the first part of the lift is [tex]h_1 = 3y[/tex]
The time taken is [tex]t_1 = 4T[/tex]
The height attained during the second part of the lift is [tex]h_2 = y[/tex]
The time taken is [tex]t_2 = T[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_1}{t_1}[/tex]
=> [tex]v_1 = \frac{3y}{4T}[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_2}{t_2}[/tex]
=> [tex]v_1 = \frac{y}{T}[/tex]
Generally the power generated during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as
[tex]F_1 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated during the second lift is
[tex]P_2 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as
[tex]F_2 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_2 = Mg * \frac{y}{T}[/tex]
So the ratio of the first power to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
This is the path which one body follows around another body in space.
Answer:
Pls ezplain your question more
Explanation:
Answer:
orbit is the answer
f F= {mango, apple, banana, orange)
Answer:
n(F) = 4Explanation:
Cardinality of a set is the number of elements in that set. Given the set.
F= {mango, apple, banana, orange), we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).
Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4
Three sticks are arranged in such a way that they form a right triangle. The
lengths of the three sticks are 0.47 m, 0.62 m, and 0.78 m. What would the
three angles of this triangle? *
Answer:
one right angel = 90 degrees
and two acute angels?
Explanation: