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vlabodo [156]
2 years ago
8

I'll give brainlest

Physics
1 answer:
musickatia [10]2 years ago
6 0

Answer:

The mouse exerts a force on the ground and the ground exerts an equal and opposite reaction force on the mouse.

At point B, Alberto's distance covered increases with time, equal distance is covered in equal time interval. The speed increases uniformly. At the point C to D, the distance covered with time remains the same. At the point D, equal distance is no longer covered in equal time intervals.

According to the law of inertia, an object will remain at rest or in a state of uniform motion unless it is acted upon by an unbalanced force. This means that smokey mouse will continue moving uniformly unless a force acts on it.

From Newton's second law; F = ma and a = F/m. F = force , m= mass, a = acceleration. This implies that smokey mouse can increase its speed by;

Increasing the force acting on the mouse

Decreasing the mass of the mouse

Action and reaction is encountered as the mouse is walking. The mouse exerts a force on the ground and the ground exerts an equal and opposite reaction force on the mouse.

Explanation:

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To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If
arlik [135]

Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

v_1^2 - 0^2 = 2(9.81)(4.0)

v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0
3 years ago
Is the star with the longest total lifetime Also the farthest from earth? Explain.
Stels [109]
E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.
6 0
3 years ago
How does this diagram demonstrate the law of superposition?
Mandarinka [93]

Answer:

well, as u can tell the top layer will always be the youngest layer aka the newest layer. The farther u go down the older the layers get. So the deeper u dig the farther back in time we see.

Explanation:

8 0
2 years ago
Read 2 more answers
A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What typ
klemol [59]

Answer:

Acceleration = 4.8 m/s²

Explanation:

Given:

Change in velocity = 19 m/s

Change in time = 4 s

Find:

Acceleration

Computation:

Acceleration = Change in velocity / Change in time

Acceleration = 19/4

Acceleration = 4.8 m/s²

Positive acceleration

8 0
3 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
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