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djverab [1.8K]
3 years ago
11

A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact

with the top of the beam, and it drives the beam 13.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Physics
1 answer:
N76 [4]3 years ago
7 0

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

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A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000
\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

3 0
3 years ago
A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont
Georgia [21]

Answer:

141.30grams

Explanation:

First, denote our known values;

J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?

Mass is impulse divided by change in velocity:

m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs

Hence, the mass of the ball is 141.30grams

3 0
3 years ago
How could you increase the kinetic energy of a wagon without increasing it's mass
Zarrin [17]

Answer:

send the wagon down a higher hill

3 0
3 years ago
Read 2 more answers
The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A
marysya [2.9K]

Answer:

option D

Explanation:

given,

A conductor is carrying current = 2.0 A is 0.5 mm thick

Hall voltage = 4.5 x 10-6 V

uniform magnetic field  =  1.2 T

density of the charge = n =?

hall voltage =V_h =\dfrac{i\ B}{n\ e\ L}

n = \dfrac{i\ B}{V\ e\ L}

n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}

n = 6.67 × 10²⁷ charges/m

hence the correct answer is option D

7 0
3 years ago
1. An object is projected upward with a velocity of 125 m/s.
grigory [225]

Answer:

A) s = 796.38 m

B) t = 12.742 s

C) T = 25.484 s

Explanation:

A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.

v = u + gt

u = 125 m/s

v = 0 m/s

g = 9.81 m/s²

Thus;

0 = 125 - 9.81(t)

g is negative because motion is against gravity. Thus;

9.81t = 125

t = 125/9.81

t = 12.742 s

Max height will be gotten from Newton's 2nd equation of motion;

s = ut + ½gt²

s = (125 × 12.742) + (½ × -9.81 × 12.742²)

s = 1592.75 - 796.37

s = 796.38 m

B) time to reach maximum height is;

t = u/g

t = 125/9.81

t = 12.742 s

C) Total time elapsed is;

T = 2u/g

T = 2 × 125/9.81

T = 25.484 s

4 0
3 years ago
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