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3 years ago

1. Four identical elephants are at different heights. If your physics teacher

Physics

1 answer:

Blababa [14]3 years ago

4 0

Elephant D does because it is at the highest height and can exert a greater amount of force leading to a greater amount of energy

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If a negatively charged particle enters a region of uniform magnetic field which is perpendicular to the particle's velocity, wi

lys-0071 [83]

Answer:

Same

Explanation:

While moving through a magnetic field in a direction perpendicular to a B-field, a continuous force experienced by a charged particle. If this magnetic field remains uniform, the force exerted also remains same and hence the velocity with which the particle is moving remains same. However, the particle is forced to move on a curved path until it forms a complete circle.

Hence, the kinetic energy remains the same because the speed is same

3 0

3 years ago

A kid pulls on a rope with 20 newtons of force. The block and tackle system pulls up a 160 newton box. What is the mechanical ad

Minchanka [31]

Answer:
The mechanical advantage of the system is 8

Explanation:
the mechanical advantage measures how much the system multiplies the input force to get the output.

In the given:
The input force (effort) is 20 Newton
The output force (load) is 160 Newton

This means that the mechanical advantage is:
mechanical advantage = load / effort = 160 / 20 = 8

Note that the mechanical advantage is unit-less (has no unit) since it is a ratio between two forces.

Hope this helps :)

3 0

3 years ago

Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi

tatiyna

If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

8 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0

3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago