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3 years ago

If we rub a rubber rod against a piece of fur the rod gains a negative change. Why does the rod have a negative charge?

Physics

1 answer:

Bingel [31]3 years ago

4 0

Answer:

because the rod takes electrons from the fur when rubbed across it

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What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).

Alex787 [66]

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

8 0

4 years ago

On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow

Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

Total heat

=10465 +167000+ 37638

=215103 J

7 0

3 years ago

Light is a form of what

mart [117]

Light is a form of energy.
(:

8 0

3 years ago

What type of issue does the effects manufacturer need

SOVA2 [1]

Answer:

A) Availability

Explanation:

Right on Edge 2021

4 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago