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2 years ago

Help please!!! Physics circular motion

Physics

1 answer:

svp [43]2 years ago

6 0

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

4. a=(v^2)/r, (don't worry about converting to meters, since units are the same in kilometers, they will cancel out) plug in values and solve.

That's it!

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How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?

soldi70 [24.7K]

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

$W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J$

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

6 0

3 years ago

What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?

sergey [27]

Answer:2.5m

Explanation:

3 0

2 years ago

for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole

sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0

3 years ago

Plz do it all i will give brainlest and thanks to best answer.

maxonik [38]

Its a, metal is a good conductor of heat so yea
Hope this helps :)

4 0

2 years ago

Read 2 more answers

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

3 years ago