7. The more friction that there is, the slower an object will move. *
true or false (science)

Answer:

70.56 J

Explanation:

Gravitational Potential Energy= mass×gravitational pull× height

= 6×9.8×1.2= 70.56 J

Answer:

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma rays. ... Microwaves are used to cook your food. Infrared waves are used in remote controls and are emitted from all warm objects, allowing them to be used to create heat-sensitive cameras.

Explanation:

brainliest please

Answer:

Distance is a scalar quantity while displacement is a vector quantity

Explanation:

A scalar quantity represents only the magnitude and does not give any detail about the direction of the quantity for example distance. Distance can be any length measured in any direction (no specific direction)

However, a vector quality represents both the magnitude and direction. For instance displacement is a vector quantity. If direction is not defined then displacement becomes equal to distance.

Total distance covered by Brandon is 18 km and total displacement covered by him is 13.41 km.

The definition of displacement is the changing of an object's position. It has a magnitude and direction and is a vector quantity. It is shown as an arrowhead that travels from the initial location to the end. An object's position changes, for instance, if it moves from position A to position B.

Distance covered by Brandon is-

12 + 6 = 18 km

Displacement covered by Brandon is-

d²= 12² + 6²

= 144 + 36

d²= 180

d = √180

d = 13.41 km.

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Answer:

Choice A. gamma rays

Explanation:

Answer:

v = 2 m/s

Explanation:

Here, we will use the law of conservation of momentum to solve this problem:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of grandma = 80 kg

m₂ = mass of James = 40 kg

u₁ = initial speed of grandma = 3 m/s

u₂ = initial speed of James = 0 m/s

v₁ = v₂ = v = final speed of grandm and James = ?

Therefore,

[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]

v = 2 m/s

Answer: e. Christian Dopplerâ

Explanation:

Based on the information given, the scientist of the past that should definitely be included in the exhibit is Christian Dopplera.

He described how the frequency of sound waves and light is being affected by the relative speed of both the source and also the observer. This was referred to as the Doppler effect.

In this scenario, the Doppler effect can be used to show how the universe is expanding. Therefore, the correct option is E.

Answer:

Explanation:

When a yellow filter is used, the stage light will appear yellow because the filter absorbs all colors of light except for yellow, which it allows through. When an orange filter is used, the stage light will appear orange for the same reason.

It is important to note that the color of the lamp itself will not change, but the light that is seen by the observer will appear to be the color of the filter due to the absorption and transmission properties of the filte

Capsule communicator or Capcom  is the only group in mission control that gets to communicate with the astronauts in space. Hence, option (B) is correct.

The capsule communicator, or Capcom, was the only voice that spoke to the astronauts during their trip to avoid any mistake. To ensure that the men in the capsule always had a familiar individual who understood their situation and could provide the information they required, Capcom was always manned by astronauts.

Both the technical control team on the ground and the astronauts in space are represented by Capcom.

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Answer:

b because we apply Hooke's law

Explanation:

Hooke's law

The displacement of the iguana between 3 s and 6 s  is 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s is  8.08 meters.

Distance is the sum of an object's movements, regardless of direction.

The term "displacement" refers to a shift in an object's position.

According to the graph:

The displacement of the iguana between 3 s and 6 s

= √{ (3-6)²+(6-0)²} meters

= 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s

= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters

= [2+ 6.08] meters

= 8.08 meters.

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The mechanical advantage of the lever is 3.5.

The ideal mechanical advantage of the lever is 4

The efficiency of the lever is 87.5 %.

The mechanical advantage of the lever is the ratio of the load overcame by the lever to the effort applied by the lever.

M.A = Load / Effort

From the graph we will find the change in load to change in effort is calculated as;

M.A = ( 160 N - 20 N ) / ( 50 N - 10 N )

M.A = 3.5

The ideal mechanical advantage of the lever is calculated as follows;

I.M.A = distance travelled by effort / distance travelled by load

I.M.A = ( 2 m ) / ( 0. 5 m )

I.M.A = 4

The efficiency of the lever is calculated as follows;

E = M.A / I.M.A x 100%

E = ( 3.5 / 4 ) x 100%

E = 87.5 %

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Answer:

d = 5 m

Explanation:

In this exercise we have a graph of displacement against time, the graph being a line, so the body has a uniform movement, the speed of the person is

         v = [tex]\frac{\Delta x}{\Delta t}[/tex]

         v =[tex]\frac{5-0}{5-0}[/tex] (5-0) / (5-0)

         v = 1 m / s

therefore the displacement is

         d = v t

         d = 1  5

          d = 5 m

Answer:

Reflected light bounces off of materials that for whatever reason cannot absorb the light. The reflected light is always whatever color it was originally. We do not see things from light that is reflected. We see things from the light that they EMIT.

Answer:

v3 = 0 gm/s / 700g

Explanation:

To solve this problem, you need to use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the total momentum of the system (the 2kg object before it explodes) is equal to the sum of the momenta of the three pieces after the explosion.

You can calculate the momentum of each piece by multiplying its mass by its velocity:

P1 = 800g * 30 m/s = 24,000 gm/s

P2 = 500g * 520 m/s = 260,000 gm/s

The total momentum of the system is the sum of these two momenta:

Ptotal = P1 + P2 = 24,000 gm/s + 260,000 gm/s = 284,000 g*m/s

The third piece has a mass of 2kg - 800g - 500g = 700g. We can use the conservation of momentum equation to find its velocity:

Ptotal = (700g * v3) + (800g * 30 m/s) + (500g * 520 m/s)

v3 = (Ptotal - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - 24,000 gm/s - 260,000 gm/s) / 700g

v3 = (284,000 - 24,000 - 260,000) gm/s / 700g

v3 = 0 gm/s / 700g

The velocity of the third mass is 0 m/s.

Hope this helps.

The frequency of the wave which remains constant as the wave propagates and the magnitude of change in amplitude is 57.2.

Density can be defined as how much matter is contained within a particular volume of a substance. A dense object is the one which weighs more than a less dense object which is the same size. An object which is less dense than that of water will float on it, whereas an object with greater density will sink. The density equation is: D = M / V.

density: dair = 1.3 x 10⁻³ g/cm³, dwater = 1.0 g/cm³,

velocity: vair = 331 m/s, vwater = 1410 m/s

v×d×A² = constant

vair×dair×Aair² = vwater×dwater×Awater²

so,

Aair/Awater = sqrt[(vwater/vair)×(dwater/dair)] = 57.2

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She might be best suited for a profession as an astronomy flight controller. The target speed for every one of the 4 motors is determined by the flight controller using the information acquired by the sensors.

The goal of astronomy is to explain whatever we see within the universe, including the planets and comets inside our own solar system, far-off galaxies, and echoes of something like the Big Bang.

Math is constantly used by astronomers. When using a telescope to see celestial objects, one application of it is this. The camera that is mounted on the telescope essentially captures a series of numbers, which could represent the amount of light that various celestial objects emit.

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Answer:

Explanation:

The mass of the ball can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

We have

[tex]m = \frac{15}{5} = 3 \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

The correct answer is - B. Increase the time it takes to stop the egg once it hits his hand.

Explanation:

In order to catch the egg the force and time makes an impulse to the egg to change its momentum and bring it to rest. In order to change the momentum, one needs to decrease the force exerted onto the egg and increase the time it takes to make it rest.

Less force can be given to the egg by increasing the time to bring it to rest stop the egg once it hits his hand. Other than this it should be caught to the sides.

the acceleration of the cat is 5m/s square

Answer:C

Explanation:

The linear expansion of the wrought iron is 0.0038 feet.

The linear expansion of the wrought iron is the increase in length of the wrought iron due to change or increase in temperature.

Mathematically, the formula for the linear expansion of metals is given as;

ΔL = αLΔθ

where;

The coefficient of thermal expansion of wrought iron = 0.0000067 / ⁰F

The linear expansion of the wrought iron is calculated as follows;

ΔL = αLΔθ

ΔL = ( 0.0000067/ ⁰F   x   8 ft    x   70 ⁰F )

ΔL = 0.0038 ft

Thus, the linear expansion of the wrought iron is a function of the increase in temperature of the iron and the original length of the iron.

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Answer:

0.5 seconds

Explanation:

Work = Fdcos(theta) = 10*2*cos(0) = 20 J

Power = W/t

40 = 20/t

t = 0.5s

The displacement, acceleration, energy and velocity of the simple harmonic motion of the mass attached to the spring are as follows;

A) x(t) = 0.05·sin(10·t + π/2)

B) The acceleration is; a(t) = -2.5 m/s²

C) The total energy is 0.0625 J

D) The velocity is ±√3/4 m/s

The restoring force of a body in simple harmonic motion is directly proportional to the displacement of the body from its mean or central position.

Mass of the block, m = 2 kg

The spring constant, k = 200 N/m

The extension of the spring = 5 cm

Time at which the spring is released, t = 0

A. The motion of the spring with the mass is a Simple Harmonic Motion

The angular velocity can be obtained using the formula;

ω = √(k/m)

Therefore;
ω = √(200/2) = 10

The angular velocity of the block on the spring is, ω = 10 rad/s

The period, T = The time to complete 2·π rad

Therefore; T = 2·π rad/(10 rad/s) = π/5 s

The amplitude, A, is the cistance of the mass from the at rest position, which is 5 cm = 0.05 m

The equation of the extension of the spring is therefore;

x(t) = 0.05·sin(10·t + c)

At t = 0, x(t) = 0.05, therefore;

sin(10 × 0 + c) = sin(c) = 1

c = π/2

The equation for the displacement as a function of time is therefore;

B. The acceleration when x(t) = A/2 is obtained as follows;

x(t) = 0.05·sin(10·t + π/2)

A/2 = 0.05·sin(10·t + π/2)

A = 0.05

0.05/2 = 0.05·sin(10·t + π/2)

sin(10·t + π/2) = 1/2

10·t + π/2 = π/6

t = -π/30

cos(10×(-π/30) + π/2) = ±√3/2

v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2) = -5 × 1/2 = -2.5

C. The energy in a pring = (1/2)·k·x²

When x = A/2, we get;

E = (1/2) × 200 × (0.05/2)² = 0.0625

D) The velocity when x = A/2 is; v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

v(t) = 0.5 × cos(10·t + π/2)

When x = A/2, sin(10·t + π/2) = 1/2, therefore;

cos(10·t + π/2) = ±√3/2

v(t) = 0.5 × ±√3/2 = ±√3/4

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The force due to gravity here is 14.7 N. The net force acting on the brick is 9.5 N and the acceleration of the brick is then 6.3 m/s².

Friction is a resistive force which opposes the motion of an object. The product of normal force by gravity and frictional coefficient gives the frictional force.

Given that, mass = 1.50 Kg

force by gravity = mg = 1.50 × 9.8 m/s² = 14.7 N

parallel component = 14.7 cos 34.4 = 12.12 N

perpendicular component  = 14.7  sin 34.4 = 8.30 N

The frictional force = mg × coefficient of friction

= 14.7 × 0.35 = 5.14 N

then net force = 14.7 - 5.14 = 9.5 N

Acceleration of the brick = net force/mass

a = 9.5 N/1.50 Kg = 6.30 m/s²

Hence, acceleration on the brick is 6.30 m/s².

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Answer:

Is at the pivot of the wheel

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V[tex]_D[/tex] = ω[tex]_c[/tex] × r[tex]_c[/tex]  ----- let this be equ 1

where V[tex]_D[/tex] is velocity of cylinder,  ω[tex]_c[/tex] is the angular velocity of drum C and r[tex]_c[/tex] is the radius of drum C

Now, Angular velocity of gear B is;

ω[tex]_B[/tex] = ω[tex]_C[/tex]

ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]  -------- let this equ 2

so;

V[tex]_D[/tex] / 0.1 m = 10V[tex]_D[/tex]

Next, we determine the angular velocity of gear A;

from the diagram;

ω[tex]_A[/tex]( 0.15 m ) = ω[tex]_B[/tex]( 0.2 m )

from equation 2; ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]

so

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / r[tex]_c[/tex] ) 0.2 m

substitutive in value of radius r[tex]_c[/tex] (0.1 m)

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / 0.1 m ) 0.2 m

ω[tex]_A[/tex]( 0.15 ) = 0.2V[tex]_D[/tex] / 0.1

ω[tex]_A[/tex] =  2V[tex]_D[/tex]  / 0.15

ω[tex]_A[/tex] = 13.333V[tex]_D[/tex]   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑[tex]U_{1-2[/tex] = T₂

0 + m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}I_A[/tex]ω²[tex]_A[/tex] + [tex]\frac{1}{2}I_B[/tex]ω²[tex]_B[/tex]

so

m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}[/tex](m[tex]_A[/tex]k[tex]_A[/tex]²)(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex](m[tex]_B[/tex]k[tex]_B[/tex]²)(10V[tex]_D[/tex])²

we given that; m[tex]_D[/tex] = 50 kg, h = 2 m, m[tex]_A[/tex] = 10 kg, k[tex]_A[/tex] 125 mm = 0.125 m, m[tex]_B[/tex] = 30 kg, k[tex]_B[/tex] = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( [tex]\frac{1}{2}[/tex] × 50 × V[tex]_D[/tex]²) + [tex]\frac{1}{2}[/tex]( 10 × (0.125)² )(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex]( 30 × (0.15)²)(10V[tex]_D[/tex])²

981 = 25V[tex]_D[/tex]² + 13.888V[tex]_D[/tex]² + 33.75V[tex]_D[/tex]²

981 = 72.638V[tex]_D[/tex]²

V[tex]_D[/tex]² = 981 / 72.638

V[tex]_D[/tex]² = 13.5053

V[tex]_D[/tex] = √13.5053

V[tex]_D[/tex] = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s