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3 years ago

A car traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the cars acceleration as it slows down?

Physics

1 answer:

mafiozo [28]3 years ago

7 0

I think I have already answer this question. Anyways please do let me know whether it is useful or not

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A 15-µF capacitor is charged to 40 V and then connected across an initially uncharged 25-µF capacitor. What is the final potenti

Kobotan [32]

Answer:

24volts

Explanation:

If a 15-µF capacitor is charged to 40V, the charge across the capacitor can be calculated using the formula;

Q = CV where;

Q is the charge flowing across the capacitor

C is the capacitance of the capacitor. = 15-µF

V is the voltage = 40V

Q = 15×10^-6×40

Q = 0.0006coulombs

If the charge of 0.0006coulombs is then connected across an initially uncharged 25-µF capacitor, the potential difference across the 25-µF can be calculated using the initial expression;

Q = CV

V = Q/C

V = 0.0006/25×10^-6

V = 24Volts

Therefore, the final potential difference across the 25-µF capacitor will be 24volts

6 0

3 years ago

Read 2 more answers

A 70 kg object strikes the ground with 2500 J of KE after falling freely from rest. How far above the ground was the object when

Kitty [74]

Answer:

3.64 m

Explanation:

m = Mass of object = 70 kg

Kinetic energy of the object = 2500 J

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

h = Height from which the object is dropped

Kinetic energy is given by

$\dfrac{1}{2}mv^2=2500\\\Rightarrow v^2=\dfrac{2\times 2500}{70}$

From conservation of energy we get kinetic energy equal to potential energy.

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{1}{2g}v^2\\\Rightarrow h=\dfrac{1}{2\times 9.81}\times \dfrac{2\times 2500}{70}\\\Rightarrow h=3.64\ \text{m}$

The object was released from a height of 3.64 m.

6 0

3 years ago

A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the

kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

$\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}$

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

$\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}$

Now for time:

$\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}$

Solving for t,

$\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}$

The total time the kangaroo takes in the air is 2.3s.

3 0

1 year ago

Mass is a fundamental physical quantity give scientific reason

Lena [83]

Answer:

yes

Explanation:

because it is independent of eachother

8 0

2 years ago

In the following situations, which ball has the greatest kinetic energy?

mixer [17]

B. A Golf ball, because although just about the same size, the mass of a golf ball is greater than a ping pong ball.

7 0

3 years ago