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3 years ago

12

What average net force is required to stop a 4.5 kg bowling ball,initially at rest, accelerated for 6 seconds over a distance of

3 m

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

32

Explanation:

4.4 x 6x3=-03o4-

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The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

3 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

2 years ago

If you pull up on a bucket with a tension force of 15 N and the bucket has a weight of 15 N, what is the net force acting on the

fredd [130]

Answer:

F = 0 [N]

Explanation:

To solve this problem we must perform a summation of forces in the direction of the vertical axis. Where the positive force is that of the tension of the upward force, while the force exerted by the weight is directed downward with a negative sign.

ΣF = 0

15 - 15 + F = 0

F = 0 [N]

3 0

3 years ago

Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let $\overset \rightarrow {v_y}$ represent the velocity of the boat across the river in the direction, $\overset \rightarrow {d_y}$ , and let, $\overset \rightarrow {v_x}$ , represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = $\overset \rightarrow {v_y}$

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is $\overset \rightarrow {v_y}$ , such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

$\overset \rightarrow {v_y}$ = 3 m/s

$\overset \rightarrow {v_x}$ = 4 m/s

The width of the river, $w_y$ = 6 meters, we have;

The resultant velocity = $\sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s$

The direction, θ, is given as follows;

$\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}$

The length of the path of the boat, <em>l</em>, is given as follows;

$l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10$

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = $\dfrac{l}{v}$ , therefore;

$t = \dfrac{10}{5} = 2$

The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

$t = \dfrac{w_y}{v_y}$

Therefore;

$t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s$

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

5 0

3 years ago

A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob

Wittaler [7]

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}$

u = 132.6 cm

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the mirror, $m=\dfrac{-v}{u}$

$m=\dfrac{-17}{132.6}$

m = -0.128

Hence, this is the required solution.

7 0

3 years ago