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Phoenix [80]
3 years ago
10

One of your fellow students comes up to you and asks the following question, "If an object is not moving, can it be accelerating

?" Based on your prior knowledge of this topic, what is the best response? A). Yes, becuase acceleration depends on the rate of change of velocity and not the value of velocity itself.
B). No, acceleration and velocity are equivalent, so if one equals zero, the other must also equal zero.
D). Yes, but only if the object's acceleration is very small.
D). Yes, but only if the acceleration is negative in order to keep velocity equal to zero.
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

correct answer is option A (Yes, becuase acceleration depends on the rate of change of velocity and not the value of velocity itself.)

Explanation:

we know that from first equation of motion,

v=u+at

0r

a=\frac{v-u}{t}............(1)

here,

v-final velocity

u-initial velocity

a-acceleration

t-time

if a body is not moving it means if its initial velocity is zero (u=0) and if after time t it moves with velocity v then its acceleration will be given as

a=\frac{v}{t}...........(2)

From above equation it is clear that if an object is not moving, it can be accelerating because acceleration depends on rate of change of velocity and not the value of velocity itself.

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If two connected points objects pass through the same set of three points, the shapes created by each will be identical, regardl
djverab [1.8K]

Answer:

True

Explanation:

It Depends on the order in which each object was plotted, if two connected points objects pass through the same set of three points, the shapes created by each point may be different

7 0
3 years ago
If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat
katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

F = k \frac{q_1q_2}{r^2}

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})

F = 3.6956*10^{-9}N

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

a_e = \frac{F}{m_e}

a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}

a_e = 4.0566*10^{21}m/s^2

The acceleration of the proton is given as,

a_p = \frac{F}{m_p}

a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}

a_p = 2.21*10^{18}m/s^2

3 0
3 years ago
A ball is thrown straight upward and rises to a maximum
Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

v_f^2 - v_i^2 = 2 a d

here we know that

v_f = 0

also we have

a = -9.81 m/s^2

so we have

0 - v_i^2 = 2(-9.81)(16)

v_i = 17.72 m/s

Now we need to find the height where its speed becomes half of initial value

so we have

v_f = 0.5 v_i

now we have

v_f^2 - v_i^2 = 2 a d

(0.5v_i)^2 - v_i^2 = 2(-9.81)h

-0.75v_i^2 = -19.62 h

0.75(17.72)^2 = 19.62 h

h = 12 m

3 0
3 years ago
An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3

Fraction of the object's weight below the surface of water is calculated as;

= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

8 0
3 years ago
Use the values provided to calculate the initial voltage
EleoNora [17]

Answer: V1= 360 V

Explanation:

7 0
3 years ago
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