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3 years ago

a bowling ball is pushed with a force of 22.0 n and acclerates at 5.5m/s2. what mass of of the bwling ball

Physics

2 answers:

Goshia [24]3 years ago

8 0

Answer:

The mass of the bowling ball is 4 kg.

Explanation:

Given that,

Force = 22.0 N

Acceleration = 5.5 m/s²

We need to calculate the mass of the bowling ball

Using newton's second law

$F = ma$

Where, F = force

m = mass of ball

a = acceleration

Put the value into the formula

$22.0=m\times5.5$

$m =\dfrac{22.0}{5.5}$

$m=4\ kg$

Hence, The mass of the bowling ball is 4 kg.

inn [45]3 years ago

4 0

A force can cause a resting object to move, or it can accelerate a moving object by changing the objects speed or direction

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3 years ago

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Emilio pushes a 100 kg freshman with 200 N of force. How much is the freshman accelerated?

ladessa [460]

Explanation:

F = MA

200 = 100 * A

A = 200/100

A = 2m/sec^2

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3 years ago

A small airplane with a wingspan of 18.0 m is flying due north at a speed of 63.6 m/s over a region where the vertical component

choli [55]

Answer:

(a) ε = 1373.8.

(b) The wingtip which is at higher potential.

Explanation:

(a) Finding the potential difference between the airplane wingtips.

Given the parameters

wingspan of the plane is = 18.0m

speed of the plane in north direction is = 70.0m/s

magnetic field of the earth is = 1.20μT

The potential difference is given as:

ε = Blv

where ε = potential difference of wingtips

B = magnetic field of earth

l = wingspan of airplane

v = speed of airplane

ε = 1.2 x 18.0 x 63.6

ε = 1373.8

(b) Which wingtip is at higher potential?

The wingtip which is at higher potential.

5 0

3 years ago

If the surface of the conductor increases 4 times what will happen to the electrical resistance?

marishachu [46]

Answer:

Increases

Explanation:

Higher current Higher resistance

Directly proportianal to each other

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3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of 6.45 m left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago