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4 years ago

A. Velocity B. Acceleration C. Time D. Displacement

Physics

1 answer:

hjlf4 years ago

6 0

I'm not too sure about this one, I've been having second guesses on it myself. But, what I put is A. Velocity. That could be wrong though, so don't take my word for it! ;)

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You are trying to push a 14 kg canoe across a beach to get it to a lake. Initially, the canoe is at rest, and you exert a force

andreyandreev [35.5K]

Answer:

I hope this is correct!

Explanation:

a) work = (force)(displacement)

We know that d = 5m, now we just need to find the force

We need to calculate the acceleration first using a = v^2 - u^2 / 2d

final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m

a = 2.1^2 - 0^2 / 2(5)

= 0.441 m/s^2

Now we can find force using f = ma

f = (14kg)(0.441m/s^2)

= 6.174 newtons

Finally, you can calculate work now that you have force!

work = (6.174n)(5m)

= 30.87 J (you may need to round sig figs!)

b) Work done by friction = (coefficient of kinetic friction)(mg)(v)

m = 14kg, g = 9.8, coefficient of friction = 0.2

force due to friction = (0.2)(14kg x 9.8)(2.1)

= 57.624 J (again you may need to round to sig figs)

c) Work you perform = total work in direction of motion + work done by friction

total work = 30.87 J, work done by friction = 57.624 J

Work you perform = 30.84 + 57.624

= 88.464 J

d) Force you applied = work you performed / distance

work you performed = 88.464 / 5 m

= 17.6929 N (again you may need to round to sig figs)

Hope this helps ya! Best of luck <3

3 0

3 years ago

Label and describe what is happening in this picture

SOVA2 [1]

Something is reproducing.

7 0

3 years ago

A 1,200 kg subcompact car accelerates at a rate of 3.0m/s

ludmilkaskok [199]

Missing question (found on internet):
"what is the force of the car?"

Solution:
According to Newton's second law, the force of the car is equal to the product between its mass and its acceleration:
$F=ma$
For the car in the problem,
$m=1200 kg$
$a=3.0 m/s^2$
So the force that accelerates the car is
$F=(1200 kg)(3.0 m/s^2)=3600 N$

6 0

3 years ago

A stationary sub uses sonar to send a 1.18x10^3 hertz sound wave through ocean water. The reflected sound wave from the flat oce

vampirchik [111]

Answer:

a) v = 1524.7 m/s

b) T = 8.47*10^-4 s

λ = 1.29 m

Explanation:

a) First, in order to calculate the speed of the sound wave, you take into account that the velocity is constant, then, you use the following formula:

$v=\frac{d}{t}$

d: distance traveled by the sound wave, which is twice the distance to the ocean bottom = 2*324 m = 648 m

t: time that sound wave takes to return to the sub = 0.425

$v=\frac{648m}{0.425s}=1524.7\frac{m}{s}$

hence, the speed of the sound wave is 1524.7 m/s

b) Next, with the value of the velocity of the wave you can calculate the wavelength of the wave, by using the following formula:

$v=\lambda f\\\\\lambda=\frac{v}{f}$

f: frequency = 1.18*10^3 Hz

$\lambda=\frac{1524.7m/s}{1.18*10^3s^{-1}}=1.29m$

And the period is:

$T=\frac{1}{f}=\frac{1}{1.18*10^3s^{-1}}=8.47*10^{-4}s$

hence, the wavelength and period of the sound wave is, respectively, 1.29m and 8.47*10^-4 s

5 0

3 years ago

Which describes an image that can be produced by a concave lens?

mixer [17]

A concave is when the bump is pressed downwards; So if the image has a downward hole, then it's concave. If it's pressed upwards like a hill or mountain, then it's convex.

5 0

4 years ago