All categories

3 years ago

8th grade science question on forces, please help me im soooo close and i dont understand this one :)

Physics

2 answers:

marin [14]3 years ago

7 0

Answer: The dog is pull the stick away from the boys hand

the boy is also pulling if the boy lets go of the stick the dog will fall due to law of action and reaction

The man is pushing the horse cart up the hill to do so he has to apply force

Explanation:

Lady_Fox [76]3 years ago

7 0

Answer:

The stick is being pulled to the right by the dog, but that force is being counter-acted by the boy, who is pulling the stick to the left.

The wagon is being pulled to the right by the horses, and the man is also pushing the cart, applying force to make it go right.

You might be interested in

PLZZZZ HELP ME if you dont know the answer plz dont help me

irina [24]

Answer: a

Explanation: genes are found in the chromeszone

6 0

3 years ago

Read 2 more answers

Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahea

vitfil [10]

Answer:

W = 819152 J = 819.15 KJ

Explanation:

The work done by Juri can be given by the following formula:

$W = FdCos\theta$

where,

W = Work done = ?

F = Force = 200 N

d = distance = 5 km = 5000 m

θ = angle to horizontal = 35°

Therefore,

W = (200 N)(5000 m)Cos 35°

5 0

3 years ago

Why don't protons stick together

3241004551 [841]

The strong nuclear force isnt quite strong enough to overcome electrosatic repulsion between two protons

4 0

3 years ago

ASAP 1-5 question plzz helppp, mark Brainliest if you answer all 5 questions

podryga [215]

Answer:

a,b,d,b, false

Explanation:

7 0

3 years ago

Read 2 more answers

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time

IrinaK [193]

Answer:

$8.0\mu C$

Explanation:

We are given that

$f=1.6 Hz$

$q=3.0\mu C=3.0\times 10^{-6} C$

$1\mu C=10^{-6} C$

Current,I= $75\mu A=75\times 10^{-6} A$

$1\mu A=10^{-6} A$

We have to find the maximum charge of the capacitor.

Charge on the capacitor, $q=q_0cos\omega t$

$\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s$

$3\times 10^{-6}=q_0cos3.2\pi t$ ....(1)

$I=\frac{dq}{dt}=-q_0\omega sin\omega t$

$75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t$ ....(2)

Equation (2) divided by equation (1)

$-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25$

$tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488$

$3.2\pi t=tan^{-1}(-2.488)=-1.188rad$

$q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C$

Hence, the maximum charge of the capacitor= $8.0\mu C$

4 0

4 years ago