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3 years ago

Long ding dong what it do?

Physics

2 answers:

Nezavi [6.7K]3 years ago

7 0

Answer:

long very very

Andre45 [30]3 years ago

3 0

Answer:

nun wbu

Explanation:

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A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.

Levart [38]

Answer:

I don't exactly know what you learned but it could be because of more friction or the bus was running out of gas.

4 0

3 years ago

Now, let's look at how the distance from the charge affects the magnitude of the electric field. Select Values on the menu, and

marin [14]

Electric field strength decreases as the distance from the source increases.

<h3>How the distance from the charge affects the magnitude of the electric field?</h3>

The strength of an electric field is inversely related to square of the distance from the source. This means that the electric field strength decreases when the distance from the source increases.

So we can conclude that Electric field strength decreases as the distance from the source increases.

Learn more about source here: brainly.com/question/25578076

#SPJ9

6 0

2 years ago

CAN YOU PLS CHECK IF ITS CORRECT I'LL MARK YOU BRAINLIST

Alexus [3.1K]

i think it looks good

yea it correct

BTW yw if it's right

5 0

3 years ago

Read 2 more answers

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t

Phoenix [80]

Answer:

a. $I=981.34 N*s$

b. $v_f=3.96 m/s$

c. $v_{f1}=3.63m/s$

d. $y_f=0.673m$

Explanation:

Given: $m=67kg$ , $h=0.720m$ , $0$

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

$I=2944-1962.66=981.35$

$I=981.34 N*s$

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$

7 0

3 years ago

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Alenkasestr [34]

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

$F=\frac{G\times m_1\times m_2}{r^2}$

G = gravitational constant

$m_1, m_2$ = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

$F\propto \frac{1}{r^2}$

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

8 0

4 years ago

Read 2 more answers