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3 years ago

1. A child sitting 1.2 m from the center of a merry-go-round moves with a speed of 1.1 m/s.

Physics

1 answer:

Ne4ueva [31]3 years ago

4 0

Answers:

A) $1m/s^{2}$

B) 22. 5 N

Explanation:

A) Since this situation is related to uniform circular motion, the centripetal acceleration $a_{C}$ of the child is calculated by:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=1.1 m/s$ is the speed

$r=1.2 m$ is the child's distane to the center (the radius)

Hence:

$a_{C}=\frac{(1.1 m/s)^{2}}{1.2 m}$

$a_{C}=1.008 m/s^{2} \approx 1 m/s^{2}$

B) Knowing th centripetal acceleration and the mass $m=22. 5 kg$ of the child, we can calculate the net horizontal force $F$ by:

$F=m a_{C}$

$F=(22. 5 kg)(1 m/s^{2})$

$F=22. 5 N$

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Write four (4) strategies on how can we manage land use around South Florida. For each strategy, explain it's benefits to the en

a_sh-v [17]

Answer:

What are some ways we can improve on our land use?

- Avoid deforestation and close the forest frontier.

The first-order priority is to end deforestation by closing the "forest frontier" or intact forests, to development.

- Increase agricultural productivity.

- Restore forests and landscapes.

- Reduce food loss and waste.

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3 0

3 years ago

You need to purchase a motor to supply 400 joules [J] in 10 seconds [s]. All of the motors you can choose from are 80% efficient

kondor19780726 [428]

Answer:

32 W.

Explanation:

Power: This can be defined as the ratio of energy to time. The S.I unit of power is watt(W). The formula for power is given as,

P = W/t.................... Equation 1

Where P = power, W = work done, t = time.

Given: W = 400 J, t = 10 s.

Substitute into equation 1

P = 400/10

P = 40 W.

If the motors that is choose is 80% efficient,

P' = P(0.8)

Where P' = minimum power

P' = 40(0.8)

P' = 32 W.

5 0

3 years ago

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$ , for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0

3 years ago

Two objects, one having twice the mass of the other, are initially at rest. Two forces, one twice as big as the other, act on th

Elodia [21]

Note: There is no image to take as a reference, so I'm assuming F2 directed to the right and F1 to the left, and F2=2F1

Answer:

$\displaystyle a=\frac{F}{3M}$

to the right

Explanation:

Net Force

When several forces are applied to a particle or a system of particles, the net force is the sum of them all, considering each force as a vector. As for the second Newton's law, the total force equals the product of the mass by the acceleration of the system:

$\vec F_n=m\cdot \vec a$

If the net force is zero, then the system of particles keeps at rest or at a constant velocity.

The system of particles described in the question consists of two objects of masses m1=M and m2, where

$m_2=2m_1=2M$

Two forces F1=F and F2 act individually on each object in opposite directions and

$F_2=2F_1=2F$

We don't get to see any image to know where the forces are applied to, so we'll assume F2 to the right and F1 to the left.

The net force of the system of particles is

$F_n=2F-F=F$

The mass of the system is

$m_t=m_1+m_2=3M$

Thus, the acceleration of the center of mass of the system is

$\displaystyle a=\frac{F}{3M}$

Since F2 is greater than F1, the direction of the acceleration is to the right.

Note: If the forces were opposite than assumed, the acceleration would be to the left

6 0

3 years ago

Help yea I need help

swat32

Answer is D I think....

8 0

3 years ago