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Y_Kistochka [10]
3 years ago
5

An object is located 51 millimeters from a diverging lens. The object has a height of 13 millimeters and the image height is 3.5

millimeters. How far in front of the lens is the image located? A. 189 millimeters B. 13.7 millimeters C. 1.12 millimeters D. 51 millimeters
Physics
2 answers:
Umnica [9.8K]3 years ago
5 0

Answer is B 13.7 millimeters


12345 [234]3 years ago
3 0
You would get 13.7 
mi/51mm=3.5mm/13mm
by solving it you will B13.7mm
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Pavlova-9 [17]
A mountain range because an ocean ridge is an underwater mountain hope this helps you 
6 0
3 years ago
Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t
sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of u_2=1\ m/s

combined velocity of the two is v=0.25\ m/s

Suppose the masses of the clouds be m_1,m_2

Conserving momentum

\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3

6 0
3 years ago
A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?
lana [24]
KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
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Hope this helps
7 0
3 years ago
Read 2 more answers
Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.
Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is t=0.00129 s

suppose u is the incoming velocity and v is the final velocity after collision

m=mass\ of\ ball

Impulse exerted is given by change in momentum of the particle.

Initial momentum P_i=m\times u

Final momentum P_f=m\times v

Change in momentum \Delta P=P_i-P_f

Impulse    J=F_{avg}\cdot t=\Delta P

J=F_{avg}\cdot t=m(u-v)

F_{avg}=\frac{m(v-u)}{t}

F_{avg}=775.2\times m(v-u)\ N                

5 0
3 years ago
An eagle is flying horizontally at a speed of 3.80 m/s when the fish in her talons wiggles loose and falls into the lake 3.90 m
Dovator [93]

Answer:

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

Explanation:

initial veetical speed V₀y=0

Horizontal speed Vx = Vx₀= 3.80m/s

Vertical drop height= 3.90m

Let Vy = vertical speed when it got to the water downward.

g= 9.81m/s² = acceleration due to gravity

From kinematics equation of motion for vertical drop

Vy²= V₀y² +2 gh

Vy²= 0 + ( 2× 9.8 × 3.90)

Vy= √76.518

Vy=8.747457

Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below

V= √Vy² + Vx²

V=√3.80² + 8.747457²

V=9.537m/s

The angle can also be calculated as

θ=tan⁻¹(Vy/Vx)

tan⁻¹( 8.747457/3.80)

=66.52⁰

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

6 0
3 years ago
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