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frozen [14]
3 years ago
13

An object with mass 1.2 kg is moving at a constant speed in a circle with radius 2.5 m. If the

Physics
1 answer:
SpyIntel [72]3 years ago
4 0

Answer:

3.9m/s² is the acceleration

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4 years ago
The scientist most often credited with the idea that matter can have wave-like properties is:
Flura [38]
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6 0
4 years ago
Read 2 more answers
A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40
Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is \frac{1}{2}mv^2

So total energy at the top of the hill =mgh+0=mgh

At the bottom of the hill kinetic energy is equal to \frac{1}{2}mv^2 and potential energy will be 0

So total energy at the bottom of the hill is equal to 0+\frac{1}{2}mv^2

Form energy conservation mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}, for v = 20 m/sec

20=\sqrt{2\times 9.8\times h}

Squaring both side

19.6h=400

h = 20.408 m

Now if velocity is 0 m/sec

40=\sqrt{2gh}

19.6h=1600

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0
3 years ago
Mechanical waves use matter to transfer energy. Please select the best answer from the choices provided T F
Semenov [28]
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8 0
4 years ago
Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you
boyakko [2]

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

v = u+at

a = \dfrac{v-u}{t}

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

a = \dfrac{21.7-0}{5.05}

a = 4.297 m/s^2

Now, using equation of motion again

For distance,

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2

s=54.79\ m

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

v_{avg}=\dfrac{D}{T}

Where, D = total distance

T = time

Put the value into the formula

v_{avg}=\dfrac{54.79}{5.05}

v_{avg}=10.85\ m/s

Hence, The distance and average speed are 54.79 m and 10.85 m.

3 0
3 years ago
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