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Flura [38]
3 years ago
10

Show work and we

Physics
1 answer:
larisa [96]3 years ago
8 0
Acceleration = final velocity - initial velocity/time taken
A = Vf-Vi/t
A = 1 m/s - 0 m/s / 4 sec
A = 1 m/s / 4 sec
A = 0.25 m/s2
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Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

\frac{v}{a_2} = 2t

You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

t_2^2 = 2t_1^2

t_2 = t_1\sqrt{2}\approx 1.14t_1

So t increased by 1.14

7 0
3 years ago
Which piece of evidence did Alfred Wegener use to develop the theory of continental drift?
Eduardwww [97]
The right answer for the question that is being asked and shown above is that: "A.tectonic activity concentrated in certain areas." A piece of evidence did Alfred Wegener use to develop the theory of continental drift is that <span>A.tectonic activity concentrated in certain areas</span>
3 0
3 years ago
A thin aluminum rod lies along the x-axis and has current of I = 16.0 A running through it in the +x-direction. The rod is in th
coldgirl [10]

Answer:

a) The magnitude of the magnetic field = 7.1 mT

b) The direction of the magnetic field is the +z direction.

Explanation:

The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by

F = (B)(I)(L) sin θ

F/L = (B)(I) sin θ

For this question,

(F/L) = 0.113 N/m

B = ?

I = 16.0 A

θ = 90°

0.113 = B × 16 × sin 90°

B = 0.113/16 = 0.0071 T = 7.1 mT

b) The direction of the magnetic field will be found using the right hand rule.

The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).

3 0
4 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
Convert 325 cm to yards I need the answer ASAP
Ira Lisetskai [31]
325 /91 = 3.571 yards
7 0
3 years ago
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