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Veronika [31]
4 years ago
8

A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t

he cars average acceleration
Physics
1 answer:
Furkat [3]4 years ago
3 0

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


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A 10-kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. what is the change in mechanica
likoan [24]

The change in mechanical energy caused by the dissipative resistance force is equal to, difference between the potential energy and kinetic energy of the object.

Potential energy of the object, P.E = mgh

m is mass of the object = 10 kg

g is acceleration due to gravity = 9.8 m/s²

h= height from which it is dropped =50 m

Substituting the value we get,

P.E = 10×9.8×50 = 4900 J

Kinetic energy of the object, K.E = \frac{1}{2}mv^{2}

v is the velocity of the object = 26 m/s²

K.E = (1/2)×10×(26)²

= 3380 J

Change in mechanical energy caused by dissipative force = P.E ₋ K.E

= 4900 ₋ 3380 = 1520 J

4 0
3 years ago
Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an
insens350 [35]

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

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

where,

F = electrostatic force of attraction or repulsion

k_e = Coulomb's constant

q_1 and q_2 are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2       ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}     ...........(2)

Equation equation 1 and 2, we get

\frac{F_{sq}}{F_{sp}}=\frac{1}{4}

F_{sq}=4\times F_{sp}

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

5 0
3 years ago
Read 2 more answers
Calculate the upthrust aciting on a body if its
Zepler [3.9K]

Answer:

As a body moving upward

T=real weight + apparent weight

T=550+490

T=1040

hope u will get the answer:)

Explanation:

7 0
3 years ago
Which statements best describe the second stage of cellular respiration? Check all that apply.
tamaranim1 [39]

Answer:

The correct answer is The stages happens in the cytoplasm.

Explanation:

5 0
3 years ago
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photon strikes the surface of tungsten and an electron is emitted. what is the maximum possible speed of the electron?
zalisa [80]

Answer: The maximum possible speed v is √2(  hν - Ф  ) / m

Explanation: You could be referring to the provided explanation, despite the fact that the question isn't comprehensive. When a photon collides with the surface of any metal, it transmits all of its energy to the electron in the atom. The collision causes the electron to travel with a certain amount of kinetic energy. This is referred to as the photoelectric effect. The maximum kinetic energy is calculated using Einstein's equation for the photoelectric effect:

K.E. = hν - Ф

½ mv² = hν - Ф

Hence the maximum possible speed is:

v = √2(  hν - Ф  ) / m

For more information on the photoelectric effect refer to this link: brainly.com/question/25027428

#SPJ4

3 0
2 years ago
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