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3 years ago

5. A backpack weighs 9.2 newtons and has a mass of 5 kg on the moon. What is the

Physics

1 answer:

Varvara68 [4.7K]3 years ago

3 0

Answer:

The strength of gravity on the moon is $1.84\ m/s^2$ .

Explanation:

We have,

Weight of the backpack, W = 9.2 N

Mass, m = 5 kg

It is required to find the strength of gravity on the moon. Weight of an object is given by :

W = mg

g is strength of gravity on the Moon

$g=\dfrac{W}{m}\\\\g=\dfrac{9.2\ N}{5\ kg}\\\\g=1.84\ m/s^2$

So, the strength of gravity on the moon is $1.84\ m/s^2$ .

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A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

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The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

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- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

$u_y=0$ is the initial vertical velocity

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Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

$v_x = 10 m/s$

Therefore, the horizontal distance covered after a time t is

$d=v_x t$

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

$d=(10)(2.02)=20 m$

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3 years ago

) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi

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Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

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Explanation:

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3 years ago

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3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

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Hence, the velocity x-component is equal to -6 m/s.

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3 years ago

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m_a_m_a [10]

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