3. A copper rod is 3 m long at a certain temperature. Calculate its length for a temperature rise of 100K

What voltage indicates the voltmeter connected to the ends of a conductor, if the ammeter connected in series with this conducto

ycow [4]

Explanation:

V= Current × Resistance

V= 400mA × 5kohms

V= 2000V

3 0

4 years ago

An action force is 40 N to the right. The reaction force must be: A. 20 N left B. 40 N left C. 20 N right D. 40 N right

Sergeeva-Olga [200]

Answer:

The answer to your question is letter B.

Explanation:

To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.

Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.

8 0

3 years ago

Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on

Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight, $w_1$ =225 N

Length,l=5 m

d=1.1 m

We have to find the distance x for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length= $\frac{l}{2}=\frac{5}{2}=2.5 m$

$r=\frac{l}{2}-d=2.5-1.1=1.4 m$

$w_2=385 N$

When system in equilibrium then

$w_1r=w_2 x$

$225\times 1.4=385\times x$

$x=\frac{225\times 1.4}{385}$

$x=0.82 m$

6 0

4 years ago

2 Points<br> What type of electromagnetic radiation is being used in the picture?

frozen [14]

Answer:

Ultraviolet radiation.

7 0

3 years ago

Read 2 more answers

A magnitic field is used to bend a beat of electrons. What uniform magnetic field is required to bend a beam of electrons moving

jarptica [38.1K]

The strength of the magnetic field is $2.73\cdot 10^{-5} T$

Explanation:

When a charged particle is moving in a uniform magnetic field, the particle experiences a force perpendicular to the direction of motion. This force is given by

$F=qvB$

where

q is the charge of the particle

v is the velocity of the particle

B is the strength of the magnetic field

Since this force acts perpendicular to the direction of motion, the particle moves in a circular motion and the force acts as a centripetal force, so we can write:

$F=qvB = m \frac{v^2}{r}$