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3 years ago

What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?

Physics

1 answer:

exis [7]3 years ago

4 0

Answer:

42.58kg

Explanation:

By Newton's second law, F = ma.

F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.

Thus, with some simple algebraic manipulation, we get the mass to equal:

m = F/a = 112N / 2.63 m/s^2 = 42.58kg

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How many significant figures are in the measurement 0.020 km?

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2 years ago

A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

nataly862011 [7]

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ( $2\pi r$ ) and the period of revolution (T), so it can be rewritten as

$v=\frac{2\pi r}{T}$

Therefore we can rewrite the acceleration as

$a=\frac{4\pi^2 r}{T^2}$

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

$T=\frac{1}{4}s = 0.25 s$

Substituting into the equation, we find the acceleration:

$a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2$

Learn more about centripetal acceleration:

brainly.com/question/2562955

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3 years ago

Two small nonconducting spheres have a total charge of 90.0 C.

valentina_108 [34]

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

8 0

3 years ago