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3 years ago

How much work is done if I use 5N of force to move an object 2 meters

Physics

1 answer:

sweet-ann [11.9K]3 years ago

8 0

It’s 10 joules. W=FD, W=5•2=10

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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago

What is the scientific method

MissTica

Answer:

Step 7- Communicate. Present/share your results. Replicate.

Step 1- Question.

Step 2-Research.

Step 3-Hypothesis.

Step 4-Experiment.

Step 5-Observations.

Step 6-Results/Conclusion

Explanation:

4 0

3 years ago

Read 2 more answers

In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a

Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m² ------------------------- (2)

from (1) and (2)

B²q²R² / m² = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0

3 years ago

What is shown in the figure above

Firdavs [7]

A single magnetic field is shown.

4 0

3 years ago