All categories

3 years ago

ASAP do all will give 25 per person pls will mark brainiest

Physics

1 answer:

neonofarm [45]3 years ago

4 0

Answer:

1) true

2) false

3) false

4) true

5) true

6) true

7) true

8) false

9) true

10) false

i think these are correct if im wrong on a few im sorry. Hope this helps at least a bit. And if i do get some wrong you know just to pick the opposite answer.

You might be interested in

If a honeybee is flying 7.0 m/s, what is the kinetic energy (Joules) if its mass is 0.1 g?

svetlana [45]

Answer:

KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules

Explanation:

mass=m=0.1g=0.0001kg

velocity=v=7.0m/s

kinetic energy=K.E.=?

as we know that

$kinetic energy=\frac{1}{2}mv^2\\ putting values\\kinetic energy=\frac{1}{2}(0.0001kg)(7.0m/s)^2\\ kinetic energy=\frac{1}{2} 0.0049joules\\kinetic energy= 0.00245Newton meter or joules$

hope it will help ^_^

5 0

3 years ago

A block is launched up a frictionless 40° slope with an initial speed v and reaches a maximum vertical height h. The same block

Strike441 [17]

Answer: 1. h

Explanation:

The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.

Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.

Try to use the formula;

1/2MV2 = mgh

Where V = √(2gh)

I hope this helps

3 0

3 years ago

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force exerted by the t

vodomira [7]

Complete Question:

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of 1.50 m and a mass of 48.0 kg. If this fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 6.30 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. (a) the salmon's acceleration m/s2 upward (b) the magnitude of the force F during this interval N

Answer:

(a) acceleration = 15.3 ms⁻²

(b) Magnitude of net force = 734.4 N

Magnitude of upward force exerted by tail fin = 1204.8 N

Explanation:

Mass of the salmon fish = 48 kg

Length of the Salmon Fish = 1.5 m

g = 9.8 ms⁻²

(a) salmon's acceleration during the time interval N:

Downward Force on the fish is equal to the Force due to gravity and is given as:

F₂ = mg

= 48 * 9.8

= 470.4 N

The direction of movement of the fish is upward and the acceleration is constant. We are given two different velocities of fish at two different instances.

- When the head breaks out of the water surface first:

Initial velocity = v₁ = 3 m/s

- When two third of its body length is out = d = 1 m

Final Velocity = v₂ = 6.3 m/s

Using the third equation of motion:

2*a*d = v₂² - v₁²

a = (6.3² - 3²)/2*1

a = 15.3 ms⁻²

(b) magnitude of force F during this interval N = ?

We are assuming that F is the net force consisting of both the upward and the downward force.

According to Newton's 2nd law of motion, Force is given as:

F = ma

F = 48 kg * 15.3

F = 734.4 N

Magnitude of upward Force = Fₓ

Force Fₓ exerted by the tail fin of the fish is given by

F = Fₓ - F₂

That is the net force is the sum of the upward and downward forces acting on the fish body. Fₓ is positive because it is in upward direction and F₂ is negative because it is in downward direction. F which is the net force here is positive as Fₓ > F₂.

=> Fₓ = F + F₂

Fₓ = 734.4 + 470.4

Fₓ = 1204.8 N

7 0

4 years ago

What is the half-life of an isotope if 125 g of a 500 g sample of the isotope remains after 3.0 years?

Troyanec [42]

Answer:

125 / 500 = 1/4 of the sample left after 3.0 yrs

1/2 * 1/2 = 1/4 sample has decayed thru 2 half-lives

2 * HL = 3

1 Half-Life = 1.5 yrs

3 0

2 years ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

3 years ago