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serg [7]
3 years ago
7

Can a particle with constant speed be accelarating?what if it has constant velocity?​

Physics
1 answer:
Olin [163]3 years ago
7 0

Explanation:

it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

When an object is moving with constant velocity, it does not change direction nor speed and therefore is represented as a straight line when graphed as distance over time.

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A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif
forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

5 0
2 years ago
Theories have both an explanatory and a predictive function. <br> a. True <br> b. False
Simora [160]
Theories have both an explanatory an a predictive function. True
3 0
3 years ago
Read 2 more answers
Determine the amount of potential energy of a 5.0Kg book that is moved to three different shelves on a bookcase. The height of e
rjkz [21]
Formula to find gravitational potential energy:
mgh
m: mass
g: gravitational acceleration
h: height (relative to reference level)

so the P.E. at 1.0.m is (5x9.8x1)= 49J
P.E. at 1.5m is (5x9.8x1.5) =73.5J
P.E. at 2.0m is (5x9.8x2)=98J
8 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
3 years ago
Unlike a physical change, a chemical change is not easily what?
krok68 [10]

because a chemical change takes time

8 0
3 years ago
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