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2 years ago

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

1 answer:

7 0

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]

= (64.2 m/s) - [ 14.7 m/s ]

= 49.5 m/s . (upward)

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8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

7 0

2 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0

2 years ago

50 J of work was performed in 20 seconds. How much power was used to do this task?

yuradex [85]

Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W

7 0

2 years ago

Read 2 more answers

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pantera1 [17]

The motion of the buoy consists of two independent motions on the horizontal and vertical axis.

On the horizontal axis, the motion of the buoy is a uniform motion with constant speed $v=50 m/s$ . On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ . The vertical position of the buoy at time t is given by
$y(t)=h- \frac{1}{2}gt^2$
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m$

8 0

2 years ago

Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Whe

joja [24]

Answer:

45.8 cm

Explanation:

To solve this, we will use the formula

5 / x² = 7/(1 - x)²

5 / x² = 7 / (1 - 2x + x²)

5 / 7 = x² / (1 - 2x + x²)

x = 0.5 * (√(35) - 5) meters

x = 0.5 * (5.916 - 5)

x = 0.5 * (0.916)

x = 0.458 or x = 45.8

8 0

2 years ago