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tamaranim1 [39]
2 years ago
8

A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box di

rectly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Physics
1 answer:
Mice21 [21]2 years ago
5 0

Answer:

Explanation:

The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .

The weight component acting on box parallel to incline plane

= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N

This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .

Force exerted by person = 98 N

distance travelled in 5 s

= velocity x time

= 2 x 5 = 10 m

Work done by person

= 98 x 10

= 980 J .

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Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

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Which of the following is the best definition of viscosity? A. the tendency of a fluid to support a floating object B. the abili
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C. example. honey has a higher viscosity than water.
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Find the magnitude of the side and the angle. Please help!
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Answer:

Explanation:

<em>Assuming the triangle is a right triangle,</em>

the magnitude of R can be found using the Pythagorean theorem,

R = sqrt(9.7^2+6^2) = 11.41 m

The angle can be found by arctangent, which is

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A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
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Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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Answer:

240km/h

Explanation:

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=240km/h

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