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storchak [24]
3 years ago
11

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x

5 20.0 cm to x 5 60.0 cm. (i) Does the electric po- tential energy of the charge-field system (a) increase, (b) re- main constant, (c) decrease, or (d) change unpredictably
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

(a) increase

Explanation:

On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.

The change in the potential energy of a charge field-system can be given as:

\delta U= -q(EdsCos \theta)

where;

q = positive test charge

E = Electric field

ds = displacement between thee charge positions

θ  = Angle between the electric field and the displacement.

Given that:

Charge of the particle = -q

displacement = (60.0 -20.0)cm = 40.0 cm

θ = 0

Replacing our values in the above equation, we have:

\delta U = -(-q)(60Cos 0)

\delta U = qEds

Since the potential energy of the system is positive, therefor the electric potential energy also increases.

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What is a substance that when mixed with water turns blue litmus paper red
alekssr [168]
When blue litmus paper is dipped in acid the paper turns red.

hope this helps :)


5 0
3 years ago
Five men push a stalled car with an average force of 400 N per person. Find the mass of the
timama [110]

The mass of the car is 2000 kg

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

In this problem, we have:

a=1 m/s^2 is the acceleration of the car

Each person applies a force of 400 N, and there are five men, so the total force applied is

\sum F = 5\cdot 400 N = 2000 N

Therefore, the mass of the car is:

m=\frac{\sum F}{a}=\frac{2000 N}{1 m/s^2}=2000 kg

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

4 0
4 years ago
Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:
FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle (v), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

v = R\cdot \omega (1)

Where:

R - Radius, measured in kilometers.

\omega - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that \omega \approx 5235.988\,\frac{rad}{s} and R = 1\times 10^{-4}\,km, then the linear velocity is:

v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)

v = 0.524\,\frac{km}{s}

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

\omega = \frac{2\pi}{T} (2)

Where T is the period, measured in seconds.

If we know that T = 31536000\,s, then the angular speed of the Earth is:

\omega = \frac{2\pi}{31536000\,s}

\omega = 1.992\times 10^{-7}\,\frac{rad}{s}

Now, we determine the linear speed:

v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)

v = 29.88\,\frac{km}{s}

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

6 0
3 years ago
PLS HELP ME!
jeyben [28]

Answer:

\underline{ \boxed{ yes}}\\

Explanation:

given : initial \: velocity \: (u )= 40 {ms}^{ - 1}  \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1}  \\ given :   - (acceleration) \: (a_r) = 2 {ms}^{ - 2}  \\ given : distance  \: (s) \: =   \: ? : \\  but \:  {v}^{2}  =  {u}^{2}  + 2( a)s\\  {0}^{2}  =  {40}^{2}  + 2( - 2)s \\  -  {40}^{2}  =  - 4s \\ s =  \frac{ -  {40}^{2} }{ - 4}  \\ s =  \frac{1600}{4}  \\s = 400 \: m

3 0
3 years ago
Which kind of thermal energy transfer warms your hand when you hold a hot
Oksanka [162]

Answer:

Conduction

Explanation:

Just took on apex

4 0
3 years ago
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