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Ivenika [448]
1 year ago
7

A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10

m/s2 and round the answer to the nearest tenth.
h1= 2m
h2=0.5m
No other information is given.
Physics
2 answers:
umka21 [38]1 year ago
8 0

The answer is: 5.5


Please mark as brainliest!

irina [24]1 year ago
5 0
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
You might be interested in
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0
1 year ago
Official (Closed) - Non Sensitive
Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

<u>t = 319.47 s</u>

5 0
2 years ago
Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe
matrenka [14]
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
6 0
1 year ago
Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to
marta [7]

Answer:

\frac{ 112.5}{15+m_{A}}=v_{f}

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed v_{iA} of 0 m/s and a mass m_{A} and the IMAX camera with an initial speed v_{ic} of 7.5 m/s and a mass m_{c} of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

3 0
1 year ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
2 years ago
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