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3 years ago

10

Physics

1 answer:

Svetlanka [38]3 years ago

4 0

Answer:

D . A mass of 5 kilograms lifted 5 meters in 10 second

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What is the string theory.

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String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.

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3 years ago

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A wire with a resistance of 20 is connected to a 12 battery. What is the current flowing through the wire?

Dima020 [189]

15.49 should be the answer if that is 12 watt battery.

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3 years ago

17. Which of the following is NOT a testable

Artemon [7]

Answer:

c. testing student opinions

Explanation:

opinions aren't factual and they would not aide an experiment if it wasn't for a social experiment.

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3 years ago

A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler

Pepsi [2]

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s

8 0

3 years ago

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

vitfil [10]

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: $F_f=\mu mg$ , backward, with

$\mu$ coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

$F-F_f = 0$

which we can solve to find the coefficient of kinetic friction:

$F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20$

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3 years ago