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4 years ago

Please help ASAP. There are 3 question it would be a big help if you can answer any.

Physics

2 answers:

sesenic [268]4 years ago

7 0

Answer:

number 1 is falling freely number 2 I think the 2nd option number 3 3rd option

erastova [34]4 years ago

7 0

1). If you're standing on a scale in an elevator, and the scale reads much less than your real weight, then the elevator must be accelerating downward.

(Think of the extreme limit: If the elevator is in free-fall, accelerating down at 9.8 m/s², then you'll apear to be weightless ... the scale will read zero.)

2). A 100 kg man is standing in an elevator, accelerating up at 2.0 m/s².

The net force acting on him is F = m·a = (100 kg) x (2.0 m/s²) = 200 N up.

The forces acting on him are:

-- gravity, m·g = (100 kg) x (9.8 m/s²) = 980 N down

-- elevator floor force . . . EF up

Net force = (980N down) + (EF up)

(200N up) = (980N down) + (EF up)

Add (980N up) to each side:

(200N up + 980N up) = EF up

Elevator force = 1180N up.

Pick the second choice: 1.2 x 10³ N up

3). My computer is resting on my desk.

There are many forces acting on my computer, and they are all balanced.

That means the NET force acting on my computer is zero.

If the net force were NOT zero, then my computer would be accelerating in the direction of the net force. But it's at rest. So the net force must be zero.

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HELP ASAP!!!!!!! PLEASE WILL GIVE BRAINLIEST

svp [43]

Answer:

cell membrane- controls what goes in and out of a cell

cell wall- protects DNA and other organelles inside a cell

ribosomes-makes and produces proteins

5 0

3 years ago

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0

4 years ago

A lens has a focal length of 0.1 m .Find the image distance when the object distance is

Black_prince [1.1K]

Answer:

A) 0.125 m (the image is real)

B) -0.4 m (the image is virtual)

Explanation:

For both situations we will use the Lens Equation:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ (1)

Where:

$f=0.1 m$ is the focal length

$u$ is the distance between the object and the lens

$v$ is the distance between the image and the lens

Isolating $v$ :

$v=\frac{uf}{u-f}}$ (2)

Knowing this, let's begin with the answers:

A) When $u=0.5 m$

$v=\frac{(0.5 m)(0.1 m)}{0.5 m-0.1 m}}$

$v=0.125 m$ The image is real

B) When $u=0.08 m$

$v=\frac{(0.08 m)(0.1 m)}{0.08 m-0.1 m}}$

$v=-0.4 m$ The negative sign indicates the image is virtual

6 0

3 years ago

A dart hits a dartboard and stops in 0.060 s. The net force on the dart is 14 N during the collision.

Rainbow [258]

Answer:

The change of momentum of the dart is 0.84 Nw.s

Explanation:

Impulse and change of momentum

The change in momentum of an object is its mass times the change in its velocity:

$\Delta p=m\Delta v=m(v_2-v_1)$

The change in the momentum can also be found by considering the force acting on it. If a force F acts for a time Δt, the change of momentum is given by:

$\Delta p=F.\Delta t$

The dart hits a dashboard with a net force of 14 N during the collision and stops in 0.06 seconds. The change of momentum is:

$\Delta p=14*0.06=0.84$

The change of momentum of the dart is 0.84 Nw.s

5 0

3 years ago

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

Brut [27]

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

8 0

4 years ago