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timama [110]
2 years ago
10

Question 12 (10 points)

Physics
1 answer:
Leni [432]2 years ago
6 0

Answer:

d jjh hhv g 7655 ijv 77_* uyfgj 3&88 huih

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2) How much work is required to pull a sled 15<br> meters if you use 30N of force?
geniusboy [140]

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Explanation:

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3 years ago
Asap pls hurry will mark brainiest
kifflom [539]
#1. A. Waxing crescent.
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3 years ago
An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and
PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

E_i+K_i+W_f=K_f+U_f

The work done by the friction force is given by:

W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]

so:

\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m

3 0
3 years ago
A toy dart gun generates a dart with a momentum of 140 kg*m/s and a
irinina [24]

Answer:

35 kg

Explanation:

From the question,

Momentum (I) = mass (m) × velocity (v)

I = m×v................... Equation 1

Where m = mass, v = velocity

make m the subject of the equation

m = I/v.................... Equation 2

Given: I = 140 kgm/s, v = 4 m/s

Substitute these values into equation 2

m = 140/4

m = 35 kg

Hence the mass of the dart is 35 kg

6 0
3 years ago
Read 2 more answers
You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

3 0
3 years ago
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