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3 years ago

"when you double the vehicle's weight, you will __________ the vehicle's stopping distance. "

1 answer:

5 0

With same braking power you will be stopping faster on the original weight therefore the answer to fill the blank is increase. The stopping distance will increase as there'll be higher energy to dissipate than lighter cars applied with the braking force similar with that of the lighter car. Also the skid and drag will add to the distance as well as the inertia of the moving heavier vehicle would be greater as well.

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A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find

garik1379 [7]

Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.

8 0

3 years ago

True or false. A convection current is the circular motion of a fluid caused by the rising of heated fluid

xz_007 [3.2K]

True that is what a convection current is

7 0

3 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago

The sport with the fastest moving ball is jai alai, where measured speeds can be 286 km/h. If a professional jai alai player fac

cricket20 [7]

Answer:

d= 794.4 cmExplanation:

Given that

Speed ,V= 286 km/h

$=286\times \dfrac{1000}{3600}\ m/s$

V=79.44 m/s

Given that time ,t= 100 ms

t= 0.1 s

We know that ( if acceleration is zero)

Distance = Speed x time

d= V t

Now by putting the values in the above equation

d = 79.44 x 0.1 m

d= 7.944 m

We know that 1 m = 100 cm

d= 794.4 cm

5 0

3 years ago

The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G

djverab [1.8K]

-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Given

A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

As the Gravity Potential energy of particle = -GMm/r

Total energy of particle = Kinetic energy + Potential Energy

As we know that

Kinetic energy = 1/2mv²

Also, v is equals to square root of GM/r

v = √GM/r

Put the value of v in the formula of kinetic energy

We get,

Kinetic Energy = GMm/2r

Total Energy = GMm/2r + (-GMm/r)

= GMm/2r - GMm/r

= -GMm/2r

Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Learn more about Gravitational Potential Energy here brainly.com/question/15896499

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3 0

1 year ago