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MrRa [10]
2 years ago
6

As a runner crosses the finish line of a race, she starts decelerating from a velocity of 9 m/s at a rate of 2 m/s^2. Take the r

unner's velocity as she crosses the finish line to be in the positive direction. What is the runner’s displacement, in meters, during the first 5 seconds after crossing the finish line?
What is her velocity, in meters per second, 5 seconds after crossing the finish line?
Physics
1 answer:
Ksivusya [100]2 years ago
4 0

Answer:

- 1 m/s, 20 m

Explanation:

u = 9 m/s, a = - 2 m/s^2, t = 5 sec

Let s be the displacement and v be the velocity after 5 seconds

Use first equation of motion.

v = u + a t

v = 9 - 2 x 5 = 9 - 10 = - 1 m/s

Use second equation of motion

s = u t + 1/2 a t^2

s = 9 x 5 - 1/2 x 2 x 5 x 5

s = 45 - 25 = 20 m

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A 1.2-m radius cylindrical region contains a uniform electric field along the cylinder axis. It is increasing uniformly with tim
eduard

Answer:

The magnitude of rate of change of electric field is 49.95\ V/m{\cdot} s.

Explanation:

Given that,

Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m

Total displacement current through a cross section of the region, I=2\times 10^{-9}\ A

We need to find the rate of change of electric field. Its is given by the formula as follows :

\dfrac{dE}{dt}=\dfrac{I}{A\epsilon_o}\\\\\dfrac{dE}{dt}=\dfrac{2\times 10^{-9}}{\pi (1.2)^2\times 8.85\times 10^{-12}}\\\\\dfrac{dE}{dt}=49.95\ V/m{\cdot} s

So, the magnitude of rate of change of electric field is 49.95\ V/m{\cdot} s.

7 0
2 years ago
the distance between the sun and earth is about 1.5X10^11 m. express this distance with an SI prefix and kilometers
Angelina_Jolie [31]
First, we write the SI prefixed. The SI unit for distance is meters.

Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²

Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga. 

1.5 x 10¹¹ /  10⁹
= 1.5 x 10² Gm or 150 Gm

Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.

1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
5 0
2 years ago
A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for
Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
2 years ago
Init. A
Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

           v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

           Δx = 20 - 15 +20

           Δx = 25 m

therefore we calculate

         v = 25/75

         v = 1/3 m / s = 0.333 m / s

in the direction of the truck

7 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
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