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2 years ago

Who can do my worksheet for me

Physics

1 answer:

timurjin [86]2 years ago

5 0

Answer:

what is it on? like name one of the questions

Explanation:

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A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position

kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

we know that,

Work = (1/2) k x²

704 = (1/2) × 500 × x²

x = 1.67 m

A spring stretched for 1.67 m distance.

4 0

2 years ago

A proton is released in a uniform electric field, and it experiences an electric force of 5 X 10^-10 N toward the South. What is

Ymorist [56]

Answer:

Electric field on proton

$E=3.12\times 10^9\ N/C$

Explanation:

Given that

$Force,F=5\times 10^{-10}\ N$

We know that

Charge on proton

$q=1.6\times 10^{-19}\ C$

We know that

Force = Electric field x Charge

F= E x q

$E=\dfrac{F}{q}\ N/C$

$E=\dfrac{5\times 10^{-10}}{1.6\times 10^{-19}}\ N/C$

$E=3.12\times 10^9\ N/C$

Electric field on proton

$E=3.12\times 10^9\ N/C$

4 0

2 years ago

A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound

Strike441 [17]

Answer:

$5.997m/s$

Explanation:

We were told to calculate the speed of the ball,

Given speed of sound as 340 m

And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.

Speed of ball = distance traveled/(time of hearing - time the sound travels).

Speed= S/t

Where S= distance traveled

t= time of hearing - time the sound travels

time=time for ball to roll+timefor sound to come back.

time of sound=16.5/340

=0.048529secs

solving for speedof ball

Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).

=16.5/(2.80-0.048529) m/s = 5.997m/s

Therefore, the speed of the ball is

5.997m/s

4 0

2 years ago

A car travels from 20-meters to 60-meters in 10 seconds. Calculate the car's speed.

slamgirl [31]

Answer:

8 m/s

Explanation:

All you have to do here is add 20 and 60 (giving you 80) and dividing by 10 seconds. 80/10= 8 m/s

3 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

2 years ago