Pada setiap tumbukan berlaku hukum

The ________ is located inside the outdoor condenser unit and receives the low-pressure refrigerant vapor through the suction li

dusya [7]

Answer:

Compressor

Explanation:

Compressor - the compressor is a unit that placed at the inner side of the condenser. The main function of the compressor is to increase the pressure by reducing the volume of air. And this reduced volume of air is compressed for the cleaning purpose of the air tube. While condenser is used for heat exchange which converts steam into liquid form.

It is advisable not to operate the compressor when the outside temperature is below 65 degrees F.

6 0

3 years ago

A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and

zzz [600]

Answer:

$F = \frac{Qq}{2\pi \epsilon_0 L d}$

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

$\lambda = \frac{q}{L}$

now the electric field due to long line charge at a distance d from it is given as

$E = \frac{2k\lambda}{d}$

$E = \frac{q}{2\pi \epsilon_0 d}$

now the force on the other charge in this electric field is given as

$F = QE$

$F = \frac{Qq}{2\pi \epsilon_0 L d}$

5 0

4 years ago

Africa's Farmers Know Best

anyanavicka [17]

1. no, they have a good method already but new tech could help in terms of laboring.

6 0

4 years ago

Please answer and ill give brainliest

11111nata11111 [884]

I think least electricity is used between probably 7-8a.m. and 4-5p.m.

This is because, around those times, the suns already out. Depending on how sunny it is, it may not be as cold as all the other times of the day. And by then, buildings are typically already warmed all up. Everybody's body heat also may play a factor in buildings. ( if there is a ton of people )

4 0

2 years ago

A batter hits a softball over a third baseman's head with speed v0 and at an angle ?from the horizontal. Immediately after the b

anastassius [24]

Answer:

a) The magnitude of the initial velocity is 18.77 m/s.

b) The launching angle is 31.51°.

c) The horizontal component of the velocity at t = 1.900 s is 16.00 m/s.

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) The horizontal component of the position vector at time t = 1.900 s is 30.40 m.

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

Explanation:

Hi there!

The equations for the velocity and position vector of the ball are the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

v = velocity vector at time t.

a and b) First, let´s find the range of the ball, i.e. the horizontal distance traveled by the ball.

The distance traveled by the baseman can be calculated with this equation:

x = v · t

Where:

x =traveled distance.

v = velocity.

t = time

Then:

x = 7.000 m/s · 2.000 s

x = 14.00 m

The baseman runs 14.00 m. Since he was located 18.00 from the home plate, the horizontal distance traveled by the ball is (14.00 m + 18.00 m) 32.00 m.

If we locate the origin of the frame of reference at the point where the ball is hit, the initial vertical and horizontal positions (x0 and y0) are zero. Since the ball is caught at the same height at which it left the bat, the vertical position of the ball when it is caught is 0.

So, the position vector of the ball at the time when it is caught (2 s after it is hit), is the following:

r = (32.00 m, 0 m)

Using the equations of the x- and y-components of the position vector, we can obtain the initial velocity and the angle:

rx = x0 + v0 · t · cos α (x0 = 0)

ry = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)

rx = 32.00 m = v0 · 2.000 s · cos α

ry = 0 m = v0 · 2.000 s · sin α - 1/2 · 9.807 m/s² · (2.000 s)²

Solving the first equation for v0:

16.00 m/s / cos α = v0

And replacing v0 in the second equation:

0 m = 32 m · sin α / cos α - 1/2 · 9.807 m/s² · (2.000 s)²

1/2 · 9.807 m/s² · (2.000 s)² = 32 m · tan α

1/2 · 9.807 m/s² · (2.000 s)² / 32 m = tan α

α = 31.51°

b) The launching angle is 31.51°

The initial velocity will be:

16.00 m/s / cos α = v0

16.00 m/s / cos (31.51°) = v0

v0 = 18.77 m/s

a) The magnitude of the initial velocity is 18.77 m/s.

c) Let´s use the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vy = v0 · sin α + g · t

The horizontal component of the velocity does not depend on time (neglecting air resistance).

Then:

vx = 18.77 m/s · cos (31.51°)

vx = 16.00 m/s

0.100 s before the ball is caught, the horizontal component of the velocity is 16.00 m/s.

Now let´s calculate the vertical component of the velocity:

vy = 18.77 m/s · sin (31.51°) - 9.807 m/s² · 1.900 s

vy = -8.823 m/s

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) Let´s use the same equations we have used in part a).

x = x0 + v0 · t · cos α

x = 18.77 m/s · 1.900 s · cos (31.51°)

x = 30.40 m

The horizontal component of the position vector at time t = 1.900 s is 30.40 m

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 18.77 m/s · 1.900 s · sin (31.51°) - 1/2 · 9.807 m/s² · (1.900 s)²

y = 0.9375 m

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

3 0

4 years ago