All categories

3 years ago

Need help ASAP please Thanks

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

3 0

Answer: D
Explanation: there is less light at that point.

You might be interested in

Which would you sprinkle around the electromagnet that you produced in the laboratory activity to see its magnet

nadya68 [22]

The Answer are:

1. iron shavings

2. compass

3. galvanometer

Give me Brainliest!!!

6 0

3 years ago

Read 2 more answers

An elephant pushes with 200 N on a load of trees. it then pushes these trees for 10 N. How much work did the elephant do?

Andreas93 [3]

Answer:

the answer is 2000Nm

Explanation:

wprk done = force × distance moved

w.d = 200N × 10m

w.d = 2000Nm

mark me as brainliest plyyzzz

7 0

4 years ago

Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the

vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

$I_{1}=1.55\ A\\I_{2}=3.15\ A$

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

$\dfrac{F}{l}=?$

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

$\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}$

where,

$\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}$

Substituting the values we get

$\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}$

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

7 0

3 years ago

Read 2 more answers

A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with

mariarad [96]

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

$F_friction=u_s*N=u_s*(mg)$

$F-F_friction=ma$

$F-u_s*(mg)=ma$

Second Newton law for the Big Block:

$F_friction=7ma$

$u_s*(mg)=7ma$

$a=u_s*g/7$

8 0

3 years ago

A car has an acceleration of -5 m/s^2. Describe the car’s motion

AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

7 0

4 years ago