SOMONE PLEASE HELP ASAP!!!!

What are some properties of transverse waves?

allsm [11]

They send out waves differently and cannot be heard easily

5 0

3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

3 years ago

When Bella pushes a box with a mass of 5.25 kilograms with a force of 15.75 newtons, it accelerates at a rate of 2.5 meters/seco

Mariana [72]

Sure !

Start with Newton's second law of motion:

   Net Force = (mass) x (acceleration) .

This formula is so useful, and so easy, that you really
should memorize it.

Now, watch:

The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?

Net Force = (mass) x (acceleration)

         = (5.25 kilograms) x (2.5 m/s²)

   Net force = 13.125 newtons .

But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?

==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.

How much ?

(Bella's 15.75 newtons) minus (13.125 that the box feels)

   = 2.625 newtons backwards, applied by friction.

5 0

3 years ago

Read 2 more answers

An observer sitting at a bus stop sees the bus drive by at 30 m/s to the east. He also sees a

Dafna1 [17]

Answer:

32 m/s

Explanation:

The speed of a bus is 30 m/s due East wrt the passenger

He also sees a passenger on the bus walking to the back at 2 m/s.

We need to find the passenger's velocity relative to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,

v = 30 m/s + 2 m/s

v = 32 m/s

Hence, the passenger's velocity relative to the bus is 32 m/s.

8 0

4 years ago