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2 years ago

• 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is

Physics

1 answer:

Kipish [7]2 years ago

6 0

Answer:

0.5 m/sec

Explanation:

v=S÷t

=5÷10

v=0.5 m/sec

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A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$

$A =\pi (0.2)^2$

$A =0.1256 m^2$

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

$\mu=IA$

$\mu=2(0.1256)$

$\mu=0.25 A\cdot m^2$

Therefore the magnetic dipole moment is $0.25A\cdot m^2$

6 0

2 years ago

A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of

PIT_PIT [208]

Answer:

Acceleration of the car will be $a=8m/sec^2$

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So $400=0\times 10+\frac{1}{2}\times a\times 10^2$

$a=8m/sec^2$

So acceleration of the car will be $a=8m/sec^2$

7 0

2 years ago

Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angul

alexandr402 [8]

Answer:

θ = 225 rad

Explanation:

given data

angle = 25 rad

to find out

angular velocity after 3t?

solution

let angular acceleration α in t

θ = ω × t + 0.5 × α × t² ........................1

here ω = 0 (initial velocity )

so put this value here

25 = 0 + 0.5 × α × t² ..........................2

α = 25 ÷ (0.5 t²)

α = 50 ÷ t² .........................3

now here we take in 3t

θ = ω × 3t + 0.5 × α × (3t)²

for ω = 0

θ = 0 + 0.5 × α × 9t²

now put value in eq 2

θ = (0.5) × (50 ÷ t²) × (3t)²

θ = 25 × 9

θ = 225 rad

6 0

2 years ago

ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is

qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = $2\pi\sqrt{\frac{l}{g} }$

where T is time period , l is length of pendulum and g is acceleration due to gravity .

$\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }$ , f is frequency of oscillation

For the given case

$\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }$

subsequently length becomes half so

$\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }$

dividing

$\frac{f}{f_o} = \sqrt{\frac{2}{1} }$

f = $\sqrt{2} f_o$

frequency of oscillation becomes √2 times.

4 0

3 years ago

1 point glitch if it works

raketka [301]

Answer:

lets see :)

Explanation:

8 0

2 years ago

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