A peregrine falcon flies at 50 m/s for 7 s. How far does it fly?

A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh

Hatshy [7]

Answer:

Explanation:

Kinetic energy of ball in motion = 1/2 m v² . Potential energy = 0

Let the minimum distance between the balls be d on collision.

Electric potential energy at that time= k Q²/d , Here kinetic energy is converted into potential energy . So

1/2 m v² = kQ²/d

d =2 k Q² / mv²,= 18 x 10⁹ x Q²/ m v².

8 0

4 years ago

Question 4

mamaluj [8]

Answer:

What is an electromagnetic wave

Explanation:

Electromagnetic radiation is also more commonly known as light, light travels in waves- and light travels at the same speed as electromagnetic waves. :)

8 0

3 years ago

Read 2 more answers

You are doing chores to earn your allowance. For each chore you do, you earn $3. What is the independent variable?

crimeas [40]

The independent variable is b the amount of chores you do

6 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an

Zolol [24]

The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
We have then:
Horizontal = 9-9.2cos (58) = 4.124742769 N.
Vertical = 9.2sin (58) = 7.802042485 N
Then, the resulting net force is:
F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
Then by definition:
F = m * a
Clearing the acceleration:
a = F / m
a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
answer:
The magnitude of the body's acceleration is
2.941756275 m / s ^ 2

6 0

3 years ago