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3 years ago

11

Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.

If you were to increase your speed to 60 miles per hour, your stopping distance is now:

2 answers:

Viefleur [7K]3 years ago

8 0

Answer:900 feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

$v^2-u^2=2as$

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph $\approx 88.002 ft/s$

$v^2-u^2=2as$

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

Rus_ich [418]3 years ago

3 0

Answer:

900 feet

Explanation:

Initial Speed, u₁ = 20 mph

Stopping distance, s₁ = 100 feet

Initial Speed, u₂ = 60 mph

Then, the stopping distance can be calculated using the third equation of motion:

$s=\frac{v^2-u^2}{2a}$

There would be same acceleration and final velocity would be zero (v=0).

$s=\frac{0-u^2}{2a}\\ \frac{s_2}{s_1}=\frac{u_2^2}{u_1^2}\\s_2= 100 ft\frac{(60)^2}{(20)^2} =900 feet$

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grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

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Then the g = F

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

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

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3 years ago

A helicopter flies 25 km north, 5 km east, then 5 km S, then 15 km W. What is the resultant displacement and direction of the he

ch4aika [34]

Answer:

Explanation:

Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)

After it flies east, the x coordinate gains 5 points, so it would now be (5,25)

After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)

After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)

East = Right

West = Left

South= Down

North = Up

I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!

Hope this helps!

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2 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

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3 years ago

Two cars are traveling in the same direction down a highway at 65 miles per hour. What is the relative velocity of the second ca

Levart [38]

Answer:

5 hours

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The cars will cover different distances because they are travelling at different speeds.

<em>D=S×T </em>

The distance travelled by the slower car = 50×x miles.

The distance travelled by the faster car = 58×x miles.

The two distances differ by 40 miles.

58x−50x=40

8x=40

x=5 hours

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A second method:

The difference in the distances is 40 miles

The difference in the speeds is #8mph.

The time to make up the 40 miles= $\frac{40}{8}$ =5 hours

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$\epsilon - Ir = RI$
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3 years ago