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liberstina [14]
2 years ago
14

Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against

each other. tom's mass is 61 kg and his little sister's mass is 15 kg. if the sister's acceleration is 2.1 m/s2 in magnitude, what is the magnitude of tom's acceleration?
Physics
1 answer:
Westkost [7]2 years ago
5 0
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".

So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>F_t = F_s
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".

From Newton's second law, we also know
</span>F=ma
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>m_t a_t = m_s a_s
<span>And find Tom's acceleration:
</span>a_t =  \frac{m_s}{m_t} a_s =  \frac{15 kg}{61 kg} (2.1 m/s^2)  =0.52 m/s^2<span>
</span>
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A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

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h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
2 years ago
A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later
mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

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8 0
1 year ago
3. In 1989, Michel Menin of France walked on a tightrope suspended under a
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Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

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Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

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Hence y = 61400/20

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Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0
2 years ago
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