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3 years ago

11

A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car doing 40 m/s. At the instant

the speeding car passes the police car, the police car accelerates from rest at 4 m/s2 to catch the speeding car. How long does it take the police car to catch the speeding car

1 answer:

eimsori [14]3 years ago

3 0

Answer:

t = 20s

Explanation:

In order for the police car to capture the other car, both cars must travel the same length.

The distance traveled by the first vehicle is:

S = ^ Vt = 40t, where t represents the elapsed time and V the vehicle speed.

The distance traveled by the police car is:

S = at ^ 2/2 = 2t ^ 2, where t represents the elapsed time, a is the acceleration of the car.

As the distances traveled are equal:

40t = 2t ^ 2 in this way t = 20s

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Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

ss7ja [257]

Answer:

The number of atoms are $1.86\times10^{8}$ .

Explanation:

Given that,

Diameter $D = 1.40\times10^{2}\ pm$

$D=1.40\times10^{2}\times10^{-12}\ m$

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

$Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}$

$Number\of\atoms =1.86\times10^{8}$

Hence, The number of atoms are $1.86\times10^{8}$ .

8 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

3 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago

The total yearly world consumption of energy is approximately 4.0 Ã— 1020 J. How much mass would have to be completely converted

BabaBlast [244]

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

$m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}$

m = 4.4 × 10³ kg

6 0

3 years ago

1 point<br> You throw a ball up in the air with a velocity of 30 m/s. How high does it<br> go?

choli [55]

Answer:

Explanation:

2as=vf^2-Vi^2

vf=30 m/s

vi= 0 m/s

a=g=9.8 m/s^2

s=vf^2-Vi^2/2a

s=(30)²-(0)²/2*9.8

s=900-0/19.6

s=45.9=46 m

4 0

2 years ago