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3 years ago

13

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

60 mph?

1 answer:

skelet666 [1.2K]3 years ago

8 0

There is more kinetic energy to stop in a 60 mph car and therefore you would have a longer stop distance.

You might be interested in

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

4 years ago

The magnetic force on a straight wire 0.69 m long is 1.5 x 10-3 N. The current in the wire is 16.9 A. What is the magnitude of t

VikaD [51]

Answer:

Magnitude of magnetic field is 1.29 x 10⁻⁴ T

Explanation:

Given :

Current flowing through the wire, I = 16.9 A

Length of wire. L = 0.69 m

Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N

Consider B be the applied magnetic field.

The relation to determine the magnetic force experienced by current carrying wire is:

F = ILBsinθ

Here θ is the angle between magnetic field and current carrying wire.

According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:

F = ILB

$B=\frac{F}{IL}$

Substitute the suitable values in the above equation.

$B=\frac{1.5\times10^{-3} }{16.9\times0.69}$

B = 1.29 x 10⁻⁴ T

7 0

4 years ago

A 4000kg truck has a head-on inelastic collision with a 2500kg truck.

iogann1982 [59]

Answer:it could be B

Explanation:

im not sure

7 0

3 years ago

How do I calculate (30 points plus a brainlyest if correct)

Nat2105 [25]

Answer:

Are you trying to calculate the net force?

If so, it would be 3 N Up.

This is because the 15 N forces from the left and right cancel out, leaving only the upwards 15 N force, and the 12 N force. However, we have to subtract 12 from 15, leaving the final net force to be 3 N Up.

Let me know if this helps!

3 0

3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

6 0

3 years ago