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Ostrovityanka [42]
3 years ago
5

Imagine that someone pushes one marble toward a motionless marble. Would there still be action-reaction forces involved in the c

ollision? How might the marbles’ motions be changed?? HELP
Physics
1 answer:
Finger [1]3 years ago
8 0
All the energy that was in the first marble was transfers into the second marble
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Which of these is an advantage of using models to study tectonic plate movements?
ratelena [41]
Actual plate movements can be made les frequent.
5 0
3 years ago
In a simple machine, the energy input is 120 J. If the efficiency of the machine is 80%, calculate the energy output
pashok25 [27]

Answer:

<h2>96 Joules</h2>

Explanation:

We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.

Solution,

Energy input of system = 120 J

Efficiency = 80% = \frac{80}{100}  = 0.8

Now,

According to definition,

Efficiency = \frac{output}{input}

Cross multiplication:

output \:  =  \: 0.8 \times 120

Calculate the product

output \:  = 96 \: joules

Hope this helps...

Good luck on your assignment...

4 0
3 years ago
Why is the unit of momentum called derived unit​
Mnenie [13.5K]

Answer:

the SI unit of momentum is :- kg.ms-1

and we know that,

kinetic energy = 1/2 mv2

E=p2/2m

p=(2Em)1/2

so the derived units are (J.kg)1/2

Explanation:

8 0
3 years ago
The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
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