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antiseptic1488 [7]
3 years ago
7

If the period of a pendulum were 2.4 s on the moon, which has an acceleration due to gravity of 1.67 m/s^2, then how long would

the rope be?
If a grasshopper can jump at 15 m/s at a 37° angle above horizontal, then how far can it Jump?

Please help asap!
Physics
1 answer:
guajiro [1.7K]3 years ago
5 0
<h2>Reffer the attachment </h2>

Mark as brainlist ❤️❤️

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What is the acceleration of this object? The object's mass is 60 kg.
andre [41]
The acceleration is 3.3 m/s2
8 0
4 years ago
Read 2 more answers
How do you find the force of a ping-pong ball rolling down a track?
Kaylis [27]

Answer:

Weight

a) weight's vertical component = Normal upward force

b) weight's horizontal component = Friction force = (mass of ball)(acceleration)

These forces depend upon the track,

1) inclined or horizontal

2) steepness.

Explanation

The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the weight which points along the direction of the ball's motion can accelerate the ball.

weight's horizontal component = Friction force = (mass of ball)(acceleration)

The other component pushes the ball into the ramp, and the ramp pushes back.

If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp. Hope this helps. Can u give me brainliest

Explanation:

3 0
3 years ago
The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
Ronch [10]

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

4 0
3 years ago
Assessment started: undefined.
mylen [45]

Answer:

conversion of momentum

Explanation:

i took the test

6 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
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