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malfutka [58]
3 years ago
6

Suppose you wanted to be able to see astronauts on the moon. What is the smallest diameter of the objective lens required to res

olve a 0.63 m object on the moon? (in m)
Possible choices:


A: 1.94
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

A:1.94

Explanation:

cause that the only one on there

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A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?
seraphim [82]

Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

=128kgm²/s²

=128 Joules

5 0
3 years ago
Read 2 more answers
A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo
Alex

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

   Speed = \frac{distance}{time taken}

   Since time is unknown, we make it the subject of the expression;

   time = \frac{distance }{speed} = \frac{400}{600}

   time = 0.67s

Learn more:

Speed brainly.com/question/10048445

#learnwithBrainly

5 0
3 years ago
At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
4 0
3 years ago
An eagle flying at 15 m/s has 600 J of kinetic energy. About how much is the eagles mass?
Eduardwww [97]

Answer:

5.33kg

Explanation:

Given parameters:

Velocity of eagle  = 15m/s

Kinetic energy of the eagle = 600J

Unknown:

Mass of the eagle  = ?

Solution:

The kinetic energy of any body is the energy due to the motion of a body. There are different forms of kinetic energy some of which are thermal, mechanical, electrical  energy.

The formula of kinetic energy is given as;

              Kinetic energy  = \frac{1}{2} m v²

where m is the mass, V is the velocity

   substitute the parameters in the equation;

                       600  = \frac{1}{2} x m x 15²

                     225m  = 1200

                            m  = \frac{1200}{225}    = 5.33kg

3 0
3 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
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