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kompoz [17]
3 years ago
9

Which of the following is a measurement of acceleration?

Physics
1 answer:
Tems11 [23]3 years ago
6 0
B is the correct answer because an object moving ten north m/s will turn into 15m/s which as you can tell is accelerating.
You might be interested in
a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
8 0
3 years ago
How does lifting electromagnet working​
dmitriy555 [2]

Explanation:

Large electrical shifting magnets have concentrated retaining strength to lift dense, ferric objects and a deep-reaching magnetization. An immensely useful materials management technique is these electromagnetic rises.

3 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
Perpetual motion machines have fascinated people especially inventors for hundreds of years before the laws of thermodynamics be
Cerrena [4.2K]

Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.

<h3>What is a perpetual motion machine?</h3>

The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.

Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.

Learn kore about a perpetual motion machine:brainly.com/question/13001849

#SPJ1

5 0
2 years ago
A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the
adelina 88 [10]

Answer:

maximum possible velocity = \sqrt{ugR}

Explanation:

centripetal acceleration when the  car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration \frac{mv^{2}}{r}

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

thereforeumg\geq \frac{mv^{2}}{R}

therefore maximum possible velocity = \sqrt{ugR}

6 0
3 years ago
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