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BigorU [14]
3 years ago
14

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone s

ubsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)

Explanation:

Time=Distance/Speed

=14.5/6.63

=2.19s(to 3s.f.)

Acceleration=Final Velocity(v)-Initial Velocity(u)/Time

9.81=v-6.63/2.19

v-6.63=21.5

v=28.1m/s

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An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by p
nexus9112 [7]

Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B  It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

        E =  \frac{\sigma}{\epsilon_o}

Where \sigma is the charge per unit area (Charge density )

From the question we are told that \sigma is doubled hence the

     E =  \frac{2 \sigma }{\epsilon_o}    

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

8 0
3 years ago
A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie
Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.  

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.  

But the work done can be written as  

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.  

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel  

each other.

 e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

               = Q_C1(T_C/T')

(T1 is the intermediate temperature)  

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 =  -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by  

e_12=1-Tc/Th

This is same as the original Carnot engine.  

7 0
3 years ago
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.60 m/s. The car is a distan
In-s [12.5K]

Answer:

101 meters

Explanation:

Distance traveled by the tourist:

d = 3.60 m/s × t

Distance traveled by the bear:

d + 39.3 m = 5.00 m/s × t

Substitute:

3.6 t + 39.3 = 5 t

39.3 = 1.4 t

t = 28.1

d = 101

7 0
3 years ago
Which statement below is true?
Veronika [31]

THe answer will be C

because if you put somethin ht next to something cold it will become cooler and if you be two things that are hot next to each other they will remain hot.

3 0
3 years ago
An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break
Anuta_ua [19.1K]

Answer:

k = 9876 N/m

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

mg(h + x) = \frac{1}{2}kx^2

also we know that maximum acceleration will be 5.3 g

so it is given as

a = \frac{k}{m} x

so we have

x = \frac{ma}{k} = \frac{5.3 mg}{k}

mg(h + \frac{5.3 mg}{k}) = \frac{1}{2}k(\frac{5.3mg}{k})^2

mg(h + \frac{5.3 mg}{k}) = \frac{14.045(mg)^2}{k}

h + \frac{5.3mg}{k} = \frac{14.045 mg}{k}

h = \frac{8.745mg}{k}

k = \frac{8.745 (1439)(9.81)}{12.5}

k = 9876 N/m

6 0
3 years ago
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