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Shtirlitz [24]
3 years ago
7

If the speed of a particle is doubled, what happens to its kinetic energy?

Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

it increases

Explanation:

maybe for example we have 5pets then you multiple them by two u get 10

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What is electromagnetic spectrum made up of
Alla [95]
The electromagnetic spectrum is made up of waves of different lengths, forming bands.
4 0
4 years ago
If vector y = 21cm and vector Z=75 cm, what is vector x ?
pychu [463]
That depends on how 'x' is related to 'y' and 'z' ... like the angles between all of them, and whether they're all on the same planet. A drawing would sure help.
6 0
4 years ago
: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature
Hunter-Best [27]

Answer:

The value is   \eta  =  0.2145  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  T_s =  100^o C =  100 + 273 = 373 \ K  

    The  second reservoir is at room temperature T_r  =  20^o C = 293 \ K

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     \eta  =  1- \frac{T_r}{T_s}

=>    1 - \frac{ 293}{373}

=>    \eta  =  0.2145

5 0
3 years ago
Your little sister (mass 25 kg) is sitting in her little red wagon (mass
Zielflug [23.3K]

Answer:

p = 60.6N*s

Explanation:

v_f = v_0+a*t

a = (v_f-v_0)/t

a = (1.8m/s)/2.35s

a = 0.77m/s²

F = m*a

F = (25kg+8.5kg)*0.77m/s²

F = 25.8N

^p = F*t

p = 25.8N*2.35s

p = 60.6N*s

5 0
3 years ago
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.
tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

4 0
3 years ago
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