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patriot [66]
3 years ago
6

A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

  Using;

                      V²  = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

     V²  = 2gH

    V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward
Anni [7]

Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0

m=18000kg

7 0
3 years ago
Under what conditions can an object have forces acting on it, but its motion does not change ?
Lady_Fox [76]

Answer: An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion.

Explanation: I hoped that helped!!

7 0
3 years ago
A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in
Kryger [21]

The average power is 3.0\cdot 10^6 W

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where :

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J

Now we can find the power required for the acceleration, which is given by

P=\frac{W}{t}

where

t = 9 s is the time elapsed

Solving:

P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th
SVETLANKA909090 [29]

Answer:

F_b  = 1647.92 N

Explanation:

Given:

length of the arm, r = 0.465 m

distance of forearm from elbow, r' = 2.15 cm = 0.0215 m

Mass of the forearm, M = 2.45 kg

Mass of the object, m = 6.55 kg

Let, the Force by bicep be, F_b

under the motionless condition, the net moment about elbow is zero

thus,

F_b × 0.0215 - Mg × (0.465/2) - mg × (0.465) = 0

or

F_b × 0.0215 - 2.45 × 9.8 × (0.465/2) - 6.55 × 9.8 × (0.465) = 0

or

F_b × 0.0215 - 5.582 - 29.848 = 0

or

F_b  = 1647.92 N

hence, the force exerted by the elbow is 1647.92 N

4 0
3 years ago
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