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MrRa [10]
3 years ago
15

A 2.0kg mass is intally moving at a speed of 10m/s due north, when a constant

Physics
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

Option(B) 56. east of north

Explanation:

<u>2D Motion </u>

If objects are moving in an XY plane subject to net force (F), acceleration (a), velocity (v) and displacement (r) in both axes, we must consider all those magnitudes as vectors because they have components in x and y.

If \vec a is the acceleration vector, then

\vec v=\vec v_o+\vec a.t

\vec r=\vec v_o.t+\frac{\vec a.t^2}{2}+\vec r_o

\vec F=m.\vec a

Assuming the positive direction to the right and upwards, we are given the following data

m=2\ kg

\vec v_o=0\ m/s\ \hat i+10\ \hat j

The x-component of the velocity is zero because it due north.

The force is applied eastward;

\vec F=10\ \hat i+0\ N\ \hat j

t=3 sec

From the formula

\vec F=m.\vec a

we can solve for \vec a

\displaystyle \vec a=\frac{\vec F}{m}

\displaystyle \vec a=\frac{10\ \hat i+0\ N\ \hat j}{2\ kg}

\displaystyle \vec a=5\ m/sec^2\ \hat i+0\ \hat j

We can now compute the velocity at t=3 sec

\vec v=0\ \hat i+10\ m/s\ \hat j+\ (5\ m/sec^2\ \hat i+0\ \hat j)3\ sec

Adding and simplifying

\vec v=15\ m/s\ \hat i+10\ m/s\ \ \hat j

The  direction is given by the angle computed as

\displaystyle tan\theta=\frac{v_y}{v_x}

\displaystyle tan\theta=\frac{10}{15}

\theta=arctan(0.667)

\theta=33.7^o\approx 34^o

This angle is north of east, the required angle is

90^o-34^o=56^o

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goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

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3 years ago
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leonid [27]
The string kind of acts like gravity 
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Paha777 [63]
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3 years ago
Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a
DaniilM [7]

Answer:

a) k = 1343.6\,\frac{N}{m}, b) l = 0.501\,m\,(50.1\,cm)

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

F \propto \Delta l

In this case, the force is equal to the weight of the object:

F = m\cdot g

F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 80.614\,N

The spring constant is:

k = \frac{F}{\Delta l}

k = \frac{80.614\,N}{0.408\,m-0.348\,m}

k = 1343.6\,\frac{N}{m}

b) The length of the spring is:

F = k\cdot (l-l_{o})

l = l_{o} + \frac{F}{k}

l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }

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6 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that
larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

T = \frac{t}{n}

T = \frac{128}{99}

T = 1.3 sec

Time period of the pendulum is also given as

T = 2\pi \sqrt{\frac{L}{g}}

1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}

g = 12.4 m/s²

4 0
3 years ago
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