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3 years ago

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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

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

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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