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KATRIN_1 [288]
3 years ago
14

Standing on a balcony, you throw your keys to a friend standing on the ground below. One second after you release the keys, they

have an instantaneous velocity of 13.9m/s, directed 45 below the horizontal. What initial velocity did you give them?
Please help me ASAP
Physics
2 answers:
GuDViN [60]3 years ago
7 0
I'm sorry I don't Know but I would add the numbers together hope that helps
Bess [88]3 years ago
5 0
There are three different equations:
Velocity=acceleration×time+ initial velocity
Position= 1/2 acceleration×time^2+initial position
Velocity^2 - initial velocity^2= 2×acceleration(position - initial position)
So you substitute or you can solve it by a system..
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barxatty [35]
Four squares are in a 2×2 grid.
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4 years ago
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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

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4 years ago
Explain how the nebular model accounts for the differences between the terrestrial and gas giant planets.
anygoal [31]
Is there any Examples/Answers or is it simply just explain it in a text box?

4 0
3 years ago
A long straight wire carries current to the right of the page. A rectangular loop is positioned directly under the wire in the p
slavikrds [6]

Answer:

Correct option A.

The net force exerted by this loop on the straight wire with the current is directed TOWARDS THE LOOP

Explanation:

The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire since typical currents consist of very large numbers of moving charges.

Given that,

The wire's current is directed towards the right of the page.

The rectangular loops carry current in a clockwise direction.

Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

Since was known that

F= iL×B

Note the current is through the wire. Then, the length is in direction of the current.

Note: this equation gives the magnetic force that acts on a length L of a straight wire carrying a current (i) and immersed in a uniform magnetic field (B), that is perpendicular to the wire.

So, the magnetic field is always perpendicular to the current.

So using right hand rule,

F = i(L×B)

The length is to the right i.e. +x direction and the Magnetic field is perpendicular to the plane, i.e. in the +z direction

F = i (L•i × B•k)

F = iLB (i×k)

F = iLB•(-j)

F = -iLB•j

Then, the force is in the negative y-direction i.e. towards the loop.

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RSB [31]

Answer:

c)both

Explanation:

The laws of reflection hold true for all the reflecting surfaces either spherical or plane. Therefore, the plane mirror, concave mirror and the convex mirror all will follow the laws of reflection.

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3 years ago
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