Ask question

Ask question

All categories

2 years ago

7

How much work is done when a porcelain vase with mass of 2.5 kg is lifted up 3 m to a shelf? (HINT- Weight can be used as Force

when the motion is up/down).

1 answer:

alexdok [17]2 years ago

8 0

Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
Work = 7.5 J
J=Jules (Jules is the unit uses to calculate work)

You might be interested in

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8

hammer [34]

Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
$W= \frac{1}{2} k (\Delta x)^2 
$
where k is the elastic constant of the spring and $\Delta x$ is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
$W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J$

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
$W_W = -F_{//} (x_2 -x_1)$
where the negative sign is given by the fact that $F_{//}$ points in the opposite direction of the displacement of the cart, and where
$F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N$
therefore, the work done by the weight is
$W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J$

8 0

2 years ago

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

mario62 [17]

Answer:

a . 0.35cm

b. 11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

$\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm$

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

$\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm$

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0

2 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

2 years ago

An airplane was moving 1200 km per minute traveling in a north-to-south direction when they were spotted by air traffic control.

Feliz [49]

Its velocity was 20 km/s southward

7 0

2 years ago

Read 2 more answers

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago