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4 years ago

PLZ PLZ PLZZZZZZZ HELP ASAP!!!

Physics

2 answers:

neonofarm [45]4 years ago

8 0

yeah it is c............................................................

AlladinOne [14]4 years ago

7 0

i belive the answe is c and if im wrong im tarrably sorry

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A 5 N force is applied to a 3 kg ball to change its velocity from 9 m/s to 3 m/s. What is the impulse on the ball ?

nika2105 [10]

So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns

3 0

3 years ago

Question 1 (1 point) There were 3 friends that decided to race their bikes down a hill without peddling. They all approached the

sattari [20]

Answer:

Explanation:

For the questions we will be using the formula;

acceleration = change in velocity/Time

acceleration = v-u/t

v is the final velocity

u is the initial velocity

t is the time

1) v = 15m/s, u = 2m/s t = 15s

a = 15-2/15

a = 13/15

a = 0.87m/s²

Their acceleration was 0.87m/s²

2) u = 10m/s, v = 45m/s, t = 1.75s

a = 45-10/1.75

a = 35/1.75

a = 35/(175/100)

a = 35×100/175

a = 3500/175

a = 20m/s²

3) u = 35m/s, v = 0m/s (comes to a stop), t = 1.25m/s

a = 0-35/1.25

a = -35/1.25

a = -28m/s²

The acceleration of the car is -28m/s²

4) v =55m/s, u = 0m/s t = 12s

a = 55-0/12

a = 55/12

a = 4.58m/s²

His acceleration when he got to the turn is 4.58m/s²

5) u = 17m/s (starting velocity)

v = 0m/s (stopping velocity)

t = 4.5s

a = 0-17/4/5

a = -17/4.5

a = -3.78m/s²

The the acceleration of the ball was -3.78m/s²

7 0

3 years ago

Two positive charges are fixed a distance apart.the sun of their charges is Qt.what charge must each have in order to maximise t

romanna [79]

Answer:

Both charges must have the same charge, Qt/2.

Explanation:

Let the two charges have charge Q1 and Q2, respectively.

Use Coulombs's Law to find an expression for the force between the two charges.

$F = k_e\frac{Q_1Q_2}{r^2}$ , where

Ke is Coulomb's contant and

r is the distance between the charges.

We know from the question that

Q1 + Q2 = Qt

So,

Q2 = Qt - Q1

$F = k_e\frac{Q_1(Q_t - Q_1)}{r^2}$

Simplify to obtain,

$F = \frac{k_e}{r^2} (Q_tQ_1 - Q_1^2)$

In order to find the value of Q1 for which F is the maximum, we will use the optimization technique of calculus.

Differentiate F with respect to Q1,

$\frac{dF}{dQ_1} = \frac{k_e}{r^2} (Q_t - 2Q_1)$

Equate the differential to 0, to obtain the value of Q1 for which F is the maximum.

$\frac{k_e}{r^2} (Q_t - 2Q_1) = 0\\Q_t - 2Q_1 = 0\\2Q_1 = Q_t\\Q1 = \frac{Q_t}{2}$

It follows that

$Q_2 = \frac{Q_t}{2}$ .

6 0

3 years ago

1. A sample of gas has a constant temperature and number of particles. As the volume of the gas sample is increased, the pressur

AURORKA [14]

Answer:

1.C

2.C

3.C

Explanation:

hope its help hehe:(

4 0

3 years ago

Can an object have both kinetic energy and gravitational potential energy? Explain.

Wewaii [24]

Answer:

Yes

Explanation:

An object can be moving (have kinetic energy) and be elevated above the ground at the same time (and also have potential energy).

3 0

3 years ago