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vitfil
8 days ago
15

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

ike nature of the board to help propel him into the air. Assume that the diver's motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. The diver is in the air for 1.7 s from when he left the board to the water. What is the speed of the diver when he reaches the water?
Express your answer with the appropriate units.
Physics
2 answers:
Keith_Richards [3.1K]8 days ago
8 0

Answer:

the final speed with which he will impact the water is v = 9.92 m/s

Explanation:

Since we know that the diver's movement primarily occurs in air,[

the diver is subject to gravitational forces and thus undergoes uniform acceleration

This allows us to utilize kinematics to establish the final velocity

v_f^2 - v_i^2 = 2ad

Consequently, the final speed at which he contacts the water is v = 9.92 m/s

Keith_Richards [3.1K]8 days ago
3 0

Answer:

10.4 m/s

Explanation:

To solve this problem, we can apply the following SUVAT equation:

v=u+at

where

v denotes the final velocity

u represents the initial velocity

a stands for acceleration

t denotes time

For the diver in our case, the values are:

the initial velocity is positive since it is directed upwards

u=+6.3 m/s

the downward acceleration from gravity is negative

a=g=-9.8 m/s^2

By substituting t = 1.7 s, we can find the velocity at the moment the diver strikes the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/sThe negative sign indicates a downward direction; hence, the diving speed will be 10.4 m/s.

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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
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Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

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The force that the car applies on the truck will be 6000 N in the reverse direction.

As per the information provided, the car's mass is one-fourth that of the truck

The truck exerts a force of 6000 N onto the car

We aim to determine the force that the car exerts on the truck

According to Newton's third law, every action has an equal and opposite reaction

Hence, the force from the car on the truck is 6000 N in the opposite direction

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An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in
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v_e/v_p = sqrt(m_p/m_e),

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The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
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A 15.0 cm object is 12.0 cm from a concave mirror that has a focal length of 4.8 cm. Its image is 8.0 cm in front of the mirror.
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The correct answer is -10.
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