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zaharov [31]
3 years ago
10

catapult’s lever holds a cannonball. The lever is attached to a tightly held rope. When the rope is released, the lever springs

forward and launches the cannonball. When the rope is held tightly, which form of energy does it possess?
Physics
1 answer:
Shalnov [3]3 years ago
6 0

Answer:

Potential energy

Explanation:

Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
Pancake syrup is an example of a
laila [671]
It is an example of liquid. if thats what you are asking for...
8 0
3 years ago
Emily wanted to see what battery lasted the longest. She put each battery into a clock and recorded the time that the clock stop
AURORKA [14]

Answer:

This question is asking to identify the following variables:

Independent variable (IV): Battery

Dependent variable (DV): Time the clock stopped

Constant: Same clock

Control: No stated control

Explanation:

The independent variable in an experiment is the variable that is subject to manipulation or change by the experimenter. In this experiment, the independent variable is the BATTERIES (Duracell, Energizer, Kroger brand, EverReady).

The dependent variable is the variable that responds to the changes made to the independent variable. It is the variable that the experimenter measures. In this case, the dependent variable is the TIME IT TAKES FOR THE CLOCK TO STOP.

Constants or control variable is the variable that the experimenter keeps constant or unchanged for all groups throughout the experiment in order not to influence the outcome of the experiment. The constant in this case is the SAME CLOCK USED.

Control group is the group that does not receive the experimental treatment or independent variable in an experiment. In this case, all groups received a different kind of battery.

6 0
3 years ago
Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?
Bond [772]

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance.  Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

7 0
3 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

brainly.com/question/14384780

#SPJ1

5 0
1 year ago
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