What amount of charge passes through a 3.0 amp television in 1.3 hours?

Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?

Roman55 [17]

Answer:

less than stating velocity due to friction and air resistance.

Explanation:

3 0

3 years ago

An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i

CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

m is the mass = 3.7 lb
g is the acceleration due to gravity = 32 ft/s^{2}
a is the acceleration of the hammer

from v = u + at

a = (v-u)/ t

a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

4 0

3 years ago

Someone places a chocolate bar near a working radar set that is used to locate ships and airplanes. Which best describes what is

Svetach [21]

Well the chocolate bar may melt at the heat of the machine but why is there a chocolate bar there in the first place is my question xD

3 0

3 years ago

a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

scoray [572]

Answer:

The answer is " $1155\ \frac{kg}{m^3}$ "

Explanation:

Please find the complete question in the attached file.

$p = p_0 + ?gh$

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

$h_1 = 11 \ cm\\\\h_2= 3 \ cm$

The pressures would be proportional to the quantity $11-3 = 8$ cm from below the surface at the interface between both the oil and the liquid.

$\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\$

$= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}$

8 0

3 years ago

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

Temka [501]

The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

8 0

3 years ago