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2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

Physics

1 answer:

noname [10]2 years ago

3 0

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

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A force pair is produced when a tennis racket strikes a tennis ball. Which of the following best explains why the tennis ball do

Serga [27]

Answer:

Each half of the force pair acts on a different object.

Explanation:

When a tennis racket strikes a tennis ball a pair force is produced. when the racket strikes the ball the racket exerts an action force on the tennis ball, according to Newton's third law for every action there is an equal and opposite reaction force, as a reaction the ball exert an equal and opposite force on the racket. These forces are often called pair forces.

As the forces acts on different bodies (Action force act on ball and reaction force act on racket) so the net force tennis ball is never zero.

4 0

2 years ago

A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis

snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

$\rho = \frac{RA}{L}$

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length of the wire

Rearranging, we have;

$R= \frac{\rho L}{A}$

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = $\pi \cdot \left (\frac{D}{2} \right )^{2} \times \frac{1}{4} = \pi \cdot \frac{D^{2}}{16} = A/4$

Therefore, the new resistance, R₂ can be expressed as follows;

$R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times \frac{\rho L}{A}$

Hence, the new resistance R₂ = 2×R, that is the resistance will be doubled.

8 0

2 years ago

A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t

Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, $a_t$ = 12.7 m/s²

mass of the truck, $m_t$ = 2490 kg

mass of the car, $m_c$ = 890 kg

let the acceleration of the car at the moment they collided = $a_c$

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

$F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2$

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0

2 years ago

How do air mass conditions ahead of the squall line support the development of new cell?

IRISSAK [1]

<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>

7 0

2 years ago

An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the

laila [671]

Answer:

a) V = 1.866 10² V , b) V = 3.424 10⁵ V , c) v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

v = 0.027 c

v = 0.027 3 10⁸

v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

Em₀ = U = q V

final point. Electrons with maximum speed

Em_f = K = ½ m v2

Em₀ = Em_{f}

e V = ½ m v²

V = ½ m v² / e

let's calculate

V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 1.866 10² V

V = 1866 V

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 3.424 10⁵ V

V = 342402 V

c) this potential difference should give the protons the same speed as the electrons

v = 8.1 10⁶ m / s

5 0

2 years ago