18. Compare the momentum of a 5450 kg truck moving at 8.00 m/s to the momentum of a 2725 kg car moving at 16.0 m/s.

The surface charge density on an infinite charged plane is −1.70 x 10^−6 C/m^2 . A proton is shot straight away from the plane a

goldfiish [28.3K]

Answer:

Acceleration of proton will be $-9.197\times 10^{12}m/sec^2$

Explanation:

We have given surface charge density of an infinite charged plate $\sigma = -1.70\times 10^{-6}C/m^2$

Electric field due to infinite sheet charge is given by $E=\frac{\sigma }{2\epsilon _0}=\frac{-1.7\times 10^{-6}}{2\times 8.85\times10^{-12}}=-0.096\times 10^6=-9.6\times 10^4N/C$

Charge in proton is given by $e=1.6\times 10^{-19}C$

So force on proton $F=qE=1.6\times 10^{-19}\times -9.6\times 10^4=-15.36\times 10^{-15}N$

Mass of proton $m=1.67\times 10^{-27}kg$

According to newtons second law force F = mass × acceleration

So $-15.36\times10^{-15}=1.67\times 10^{-27}a$

$a=-9.197\times 10^{12}m/sec^2$

3 0

3 years ago

Which is a characteristic of the image formed between F and the center of the lens?

m_a_m_a [10]

Answer: The image is upright

Explanation:

The lens shown in the diagram is concave lens. It is a diverging lens which means that the light rays which fall from the object onto the lens diverge.

Here, the object is kept between focus and center of lens. The image would form between focus and center of the lens at the same side of the object. The image would virtual, diminished and upright. Thus, the correct option is: The image is upright.

3 0

3 years ago

A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end o

Olegator [25]

Answer:

Explanation:

Given

Weight of strut $W_1=400 N$

Weight of sign Board $W_2=200 N$

In this question angle which cable makes with horizontal is not given so

assuming $\theta =30^{\circ}$

From Diagram

Moment about hinge point

$T\sin 30\times L=W_1\times \frac{L}{2}+W_2\times L$

$T\sin 30=\frac{W_1}{2}+W_2$

$T\sin 30=\frac{400}{2}+200$

$T=800 N$

$\sum F$ in x direction is zero

$F_x=T\cos 30$

$F_x=800\cos 30=346.41 N$

$\sum F$ in Y direction is zero therefore

$F_y+T\sin 30=W_1+W_2$

$F_y=400+200-800\cdot \sin 30$

$F_y=200 N$

$F_{net}\ at\ hinge=\sqrt{F_x^2+F_y^2}$

$F_{net}=399.99\approx 400 N$

6 0

3 years ago

What is the magnification of a virtual image if the image is 10.0 cm from a lens and the object is 30 cm from the lens

7nadin3 [17]

Answer:

The magnification of a virtual image is -0.33

Explanation:

We can calculate the magnification as the ratio of height of the image to the height of the object. It can be also stated as the ratio of size of the image obtained to the original size of the object. Also this ratio can be equated to the ratio of image distance to object distance. So magnification can be termed as ratio of image distance to object distance. As in the present case, the image distance or the distance of the formation of image from the lens is 10 cm and the distance of object from the lens or the object distance is 30 cm. Then, $Magnification =-\frac{Image distance}{Object distance}=-\frac{10}{30}=-0.33$ .

So, the image will be diminished to 0.33 times the original size as the magnification of the virtual image is obtained as -0.33.

5 0

3 years ago

Read 2 more answers

balandron [24]

Idk characteristics

6 0

3 years ago

18. Compare the momentum of a 5450 kg truck moving at 8.00 m/s to the momentum of a 2725 kg car moving at 16.0 m/s.​

18. Compare the momentum of a 5450 kg truck moving at 8.00 m/s to the momentum of a 2725 kg car moving at 16.0 m/s.