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2 years ago

Rachel and Sarah are on a bus travelling at 5 mph past John who is standing on the sidewalk. Rachel then throws a ball

Physics

1 answer:

oksian1 [2.3K]2 years ago

5 0

Answer:

I think C

Explanation:

Since the bus is moving away from John.

{C - V}.

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A notebook is on a desk the force of the notebook weight pushing down on the table is 10 in the force of the table normal force

Mamont248 [21]

Answer:

Yes it is balanced

Explanation:

Because 10-10= a net force of 0N

Its equal (balanced)

5 0

3 years ago

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v

lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

3 0

3 years ago

It is important that your muscles are very warm when doing this type of stretching.

MaRussiya [10]

Answer:

ballistic stretching

3 0

3 years ago

Read 2 more answers

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

3 years ago

Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

$F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g$

Also, the car is making a circular motion:

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

Combining the equations:

$\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}$

Finally the angle is

$\arcsin(v^2/(gR))$

4 0

3 years ago