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aksik [14]
2 years ago
14

A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr

op in the next 6.6 s
Physics
1 answer:
jeka942 years ago
6 0

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

h = ut + 1/2 g t²

u again = 0

h = .5 x 9.8 x 9.9²

= 480.24 m

distance travelled in last 6.6 s

= 480.24 - 53.4

= 426.84 m

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If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0 ∘C to 85.0 ∘C ?
andrey2020 [161]

Answer:

Explanation:

C_{water} = 4190 J/kg.K

C_{Al} = 910 J/Kg. K

m_{Al} = 1.50 kg

m_{water} = 1.80 kg

Q_{added} = Q_{Al} + Q_{water}

=m_{Al} C_{Al}ΔT + m_{water} C_{water}ΔT

= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)

= 578,955 J

= 579 kJ

3 0
3 years ago
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Mila [183]

Answer:

v=6.65m/sec

Explanation:

From the Question we are told that:

Mass m=97.6

Coefficient of kinetic friction  \mu k=0.555

Generally the equation for Frictional force is mathematically given by

 F=\mu mg

 F=0.555*97.6*9.8

 F=531.388N

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 a_f=-\mu g

 a_f=-0.555 *9.81

 a_f=-54455m/sec^2

Therefore

 v=u-at

 v=0+5.45*1.22

 v=6.65m/sec

4 0
3 years ago
Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa
rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

\theta =1.22\cdot \frac{\lambda }{D}

Applying the given values we get

\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads

thus the linear separation equals L=\theta \times Distance

Applying the given values we get

L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes

6 0
3 years ago
When conducting scientific research, any information that is found must be analyzed to determine its validity. John believes tha
SashulF [63]

Answer:

b. "One thing for sure is that they were UFOs and they had to be piloted by some type of intelligent life."

Explanation:

Well...generally the above observation may seem wrong but it isn't. That's because according to scientific point of view, if he saw a real UFO flying out in the air then it is logically possible that some being of an advance civilization must be piloting that craft. If the observation is correct then it can't just be a coincidence at all.

5 0
3 years ago
Read 2 more answers
If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.
Schach [20]

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

6 0
3 years ago
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