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2 years ago

Why does the moon appear in the daytime

Physics

1 answer:

Brut [27]2 years ago

4 0

Answer:

You can see the moon when it is in the perfect spot and it's reflecting the right amount of light. :)

Hope this helps! If you don't mind, please mark as brainliest! Thx!

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your softball coach needs to add a new player to the team roster. which of these people would most likely demonstrate the best s

Shkiper50 [21]

It would be easier to answer your question if you attached options. Anyway I remember that the right answer to that question is:<span> marissa, because she has cheered for the team all season long</span>

6 0

3 years ago

One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0

N = $W_m +w_{table}$

N = $M_m g + w_{table}$

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

N = 65 9.8 + 100

N = 737 N

c) To calculate the mass of the table we use the relation

W = $m_{table}$ g

$m_{table} = \frac{w_{table}}{g}$

$m_{tabble}= \frac{100}{9.8}$

$m_{table}$ e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Learn more here: brainly.com/question/19860811

7 0

2 years ago

A cobalt atom contains 27 protons, 32 neutrons, and 27 electrons. What is

kirza4 [7]

32? I could be wrong but I’m going with that answer choice

3 0

2 years ago

You are tossing a ball directly up into the air. Before the ball leaves your hand, you are exerting a force directed against the

DaniilM [7]

Answer:

F n = 0.2 N

Explanation:

given,

you are exerting force of 10 N on the ball.

mass of the ball = 1 kg

acceleration due to gravity = 9.8 m/s²

normal force on the ball = ?

normal force is force exerted by the object to counteract the force from other object.

normal force acting on the ball will be

F n = F - mg

F n = 10 - 1 × 9.8

F n = 10 -9.8

F n = 0.2 N

Hence, normal force acting on the ball is equal to 0.2 N

7 0

3 years ago

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

<u>Note that the acceleration is negative because is a braking process.</u>

4 0

3 years ago

Read 2 more answers