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Mama L [17]
3 years ago
12

The driver of a car slams on the brakes, causing the car to slow down at a rate of 17ft/s2 as the car skids 285ft to a stop.

Physics
1 answer:
ozzi3 years ago
8 0

1) 5.79 s

2) 98.4 ft/s

Explanation:

1)

The motion of the car is a uniformly accelerated motion (it means it travels with constant acceleration), so we can find the time it takes for the car to stop by using the following suvat equation:

s=vt-\frac{1}{2}at^2

where

s is the distance travelled

v is the final velocity

t is the time

a is the acceleration of the car

In this problem we have:

s = 285 ft is the distance travelled

a=-17 ft/s^2 is the acceleration of the car (negative since the car is slowing down)

v = 0 ft/s is the final velocity of the car, since it comes to a stop

Solving for t, we find:

t=\sqrt{\frac{-2s}{a}}=\sqrt{\frac{-2(285)}{-17}}=5.79 s

2)

The initial speed of the car can be found by using another suvat equation, namely:

v=u+at

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In this problem, we have:

v = 0 is the final speed of the car

a=-17 ft/s^2 is the acceleration of the car (negative since the car is slowing down)

t = 5.79 s is the total time of motion (found in part 1)

Therefore, the initial speed of the car is:

u=v-at=0-(-17)(5.79)=98.4 ft/s

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A spacecraft travels at 1.5 X 108 m/s relative to Earth. A process onboard the
kvasek [131]

Answer:

73.6 minutes

Explanation:

relative time = time interval / √(1 - observer velocity² / speed of light²)

we have relative time. we want time interval.

rearrange

time interval = relative time x √(1 - observer velocity² / speed of light²)

convert 85 mins into seconds

85 x 60 = 5100

1.5 x 10⁸ as a number is 150000000

for c = 299 792 458

time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)

for c = 3 x 10⁸

time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)

time interval = 5100 x 0.866

time interval = 4415.71

divide by 60 for back into minutes

time = 73.6 minutes

4 0
2 years ago
Once a scientist has made a hypothesis, what would they typically do next? (2 points)
ipn [44]

Answer:

b-testing

Explanation:

First would be observation/research. Then the hypothesis. After that you would test your theory, conduct experiments. And finally, your conclusion- what you got from the whole process basically.

Hope this helps.

4 0
3 years ago
As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
marshall27 [118]

Answer:

The resultant velocity is  v_t=10 knots

Explanation:

Apply the law of conservation of momentum

     M_L *v_L + M_f * V_f = (M_L + M_f) v_t

Where M_L is the mass of the Luxury Liner = 40,000 ton

            v_L is the velocity of Luxury Liner = 20 knots due west

            M_f mass of freighter = 60,000

           v_f is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

         v_f = 0 \ knots

So the equation would be

              M_L *v_L = (M_L + M_f) v_t

Substituting values

            40000*20 = (40000+ 60000)v_t_w

Where v_t_w the final velocity due west

Making v_t_w the subject

          v_t_w = \frac{40,000* 20}{(40000 + 60000)}

                = 8 \ knots

Apply the law of conservation of momentum toward the the north direction          

          v_L = 0 \ knots

So the equation would be

           M_f *v_f = (M_L + M_f) v_t_n

Where v_t_n the final velocity due north

     Making v_t_n the subject

          v_t_n = \frac{60,000* 10}{(40000 + 60000)}

                = 6 \ knots

The resultant velocity is

       v_t = \sqrt{v_t_w^2 + v_t_n^2}

            = \sqrt{8^2 +6^2}

           v_t=10 knots

8 0
3 years ago
For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical
nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.


If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.


If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.


If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0
2 years ago
Read 2 more answers
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