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irinina [24]
3 years ago
6

Help me again...(science)

Physics
1 answer:
Lapatulllka [165]3 years ago
8 0
The first question would be A
The second would be either A or D
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Please help me with 1&2
Oxana [17]

Answer:

1: A

Explanation:

7 0
3 years ago
An object at 20∘c absorbs 25.0 j of heat. what is the change in entropy δs of the object?
anastassius [24]
From the definition of entropy, the entropy change of an object is
\delta S =  \frac{Q}{T}
where
Q is the heat absorbed
T is the absolute temperature

in our problem, we have Q=25.0 J, while the absolute temperature is (converting in Kelvin)
T=20 ^{\circ}C + 273 = 293 K

and so the entropy change is
\delta S=  \frac{25.0 J}{293 K}=0.085 JK^{-1}
5 0
4 years ago
Read 2 more answers
Young's experiment is performed with light of wavelength 502nm from excited helium atoms. Fringes are measured carefully on a sc
alexandr402 [8]

Answer:

d = 1.30 mm

Explanation:

given,

wavelength of the light source (λ)= 502 nm

slits is separated (d) = ?

distance to form interference pattern(D) = 1.35 m  

20 th fringe

y = 10.4 mm = 0.0104 m

now,separation between to slits

 d = \dfrac{m\lambda\ D}{y}

where y is the fringe width

 d = \dfrac{20 \times 502\times 10^{-9}\ \times 1.35}{0.0104}

 d = \dfrac{13354\times 10^{-9}}{0.0104}

      d = 1.30 mm

separation between to slits is equal to d = 1.30 mm

8 0
3 years ago
What is gama rays an it's uses​
pav-90 [236]

Answer:

GAMMA RAYS:

A photon emitted spontaneously by a radioactive substance also : a photon of higher energy than that of an X-ray.

USES OF GAMMA RAYS:

Gamma rays are used in medicine (radiotherapy), industry (sterilization and disinfection), and the nuclear industry. Shielding against gamma rays is essential because they can cause diseases to skin or blood, eye disorders, and cancers.

5 0
3 years ago
A ball is thrown from the top of a building with an initial velocity of 21.9 m/s straight upward, at an initial height of 51.6 m
nalin [4]

Part A)

when ball will reach to highest point then it's speed will become zero

so we can use kinematics to find the time

v_f = v_i + at

0 = 21.9 + (-9.8) t

0 = 21.9 - 9.8 t

t = 2.23 s

Part b)

for finding the maximum height we can use another kinematics equation

v_f^2 - v_i^2 = 2ad

0 - 21.9^2 = 2(-9.8)(H)

H = \frac{21.9^2}{19.6} = 1.12 m

so it will rise to 1.12 m from the point of projection

Part C)

Ball will take double the time which it take to reach the top point.

So here the time to reach the top is 2.23 s

so time taken by the ball to reach at same point after projection is given as

t = 2(2.23) = 4.46 s

Since ball have reached to same point so the final velocity must be same as initial velocity

so we have

v_f = 21.9 m/s downwards

Part d)

when ball reached to the bottom

displacement of ball = -51.6 m

a = -9.8 m/s^2

v_i = 21.9 m/s

now by kinematics we have

d = v_i t + \frac{1}{2}at^2

-51.6 = (21.9)t + \frac{1}{2}(-9.8)t^2

4.9 t^2 - 21.9 t - 51.6 = 0

by solving above equation we have

t = 6.2 s

now for the velocity at that instant we have

v_f = v_i + at

v_f = 21.9 - (9.8) (6.2)

v_f = -38.6 m/s

so its velocity is 38.6 m/s downwards

Part e)

for the position of ball at t = 5.35 s we can use

d = v_i t + \frac{1}{2}at^2

d = 21.9(5.35) + \frac{1}{2}(-9.8)(5.35)^2

d = -23.1 m

so it is 23.1 m below the initial position from which it is thrown

now for the velocity we can say

v_f = v_i + at

v_f = 21.9 + (-9.8)(5.35)

v_f = -30.53 m/s

so it will be moving downwards with speed 30.53 m/s

8 0
3 years ago
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