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marta [7]
3 years ago
14

Please help this due 11:59

Physics
2 answers:
melamori03 [73]3 years ago
5 0

Answer:

it's x that is the right answer

zlopas [31]3 years ago
5 0
The correct answer is Z mate
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bulgar [2K]

Answer:

the answer is the temperatures of both beakers' water will increase by the same amount...

Explanation:

I know this because i just did it on study island

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3 years ago
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How does momentum relate to impulse ?
fomenos

Answer:

C. Impulse = F*t=(m*a)*t= m*(a*t) = m*Dv= D(Momentum) (“D” here’s mean Delta so change in)

Explanation:

In fact, the impulse is equal to the change in momentum of an object.

Impulse is defined as the product between the force (F) and the time (t):

I=Ft

however, the force is defined as the product between mass (m) and acceleration (a):

F=ma\\I=(ma)t

But the product a (acceleration) times t (time) is equal to the change in velocity of the object:

I=m(at)=m \Delta v

And this is exactly the definition of change in momentum:

I = m\Delta v = \Delta p

5 0
3 years ago
How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?
mario62 [17]

Answer:

Required time, t = 0.84 seconds

Explanation:

It is given that,

Initial speed of an object, u = 22 m/s

Final velocity of an object, v = 27 m/s

Acceleration, a = 5.93 m/s²

We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:

v=u+at

Where

t = time

t=\dfrac{v-u}{a}

t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}

t = 0.84 seconds

Hence, the time required for an object is 0.84 seconds.

4 0
3 years ago
A particle moves in a straight line so that time t seconds its displacement is x metres from a fixed point is given by x=3t^2 +2
Andru [333]

Answer:

-5 meters

Explanation:

x = 3t² + 2t − 5

The initial displacement is when t = 0.

x = 3(0)² + 2(0) − 5

x = -5

6 0
3 years ago
What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?
Nuetrik [128]
The potential at a distance r from a charge Q is given by
V(r) = k_e  \frac{Q}{r}
where ke is the Coulomb's constant.

The charge in our problem is Q=2.9 nC=2.9 \cdot 10^{-9} C; for the point at r=0.48 mm=0.48 \cdot 10^{-3} m, the potential is
V_1 = k_e  \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}=  5.43 \cdot 10^4 V

For the point at infinity, we immediately see that the potential is zero, because r= \infty and so V_2 = 0.

Therefore, the potential difference between the two points is
\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V
6 0
3 years ago
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