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2 years ago

Name the quantity which is measured by the area occupied below the velocity time graph

2 answers:

3 0

If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.

dmitriy555 [2]2 years ago

3 0

If you're graphing velocity vs. time, you're denoting what velocity you're moving at various points at one time

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A metal ring 4.30 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

CaHeK987 [17]

Answer:

A)0.00966 N/C

B) counterclockwise direction

Explanation:

We are given;

Diameter of the metal ring; d = 4.3 cm

Radius;r = 2.15 cm = 0.021- m

Initial magnetic field, B = 1.12 T

Rate of decrease of the magnetic field;dB/dt = 0.23 T/s

Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

We are given;

E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

6 0

2 years ago

What type of machine is used to measure the location of an earthquake epicenter.

Ber [7]

Answer:

Seismometer

Explanation:

Seismometer measures Earth quake epicenter while seismograph measures Earth quake size

8 0

1 year ago

An arrow in flight has an initial velocity of 65 meters per second, and 10 seconds later, it has a velocity of 35 meters per sec

-BARSIC- [3]

Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2

7 0

3 years ago

A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; $m_{total$ v

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0

2 years ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

2 years ago