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3 years ago

Acceleration equation physics

Physics

2 answers:

ella [17]3 years ago

7 0

Answer:

F=ma

dv/dt= a

a is referred to as acceleration

Rudiy273 years ago

7 0

Answer:

Acceleration=change in velocity ➗ time

Explanation:

Acceleration=change in velocity ➗ time

Whereby:

Change in velocity=final velocity - initial velocity

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The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg

nikitadnepr [17]

By applying Newton's second law of motion;

ma = mg - T

Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.

For the current case, the velocity is constant therefore,
a = 0

Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N

Tension in the cable is 1128.15 N.

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3 years ago

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Stella [2.4K]

Answer:

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Explanation:

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3 years ago

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if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

Paul [167]

The concepts of force, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops pushing? ... F net =10 N 2 N. = 8 N (to the right) a = F net m. = 8 N. 5 kg. =1.6 m s. 2 ... Two equal forces act on an object in the directions shown. If these ... Two connected carts being accelerated by a force F applied by.

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3 years ago

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

kotykmax [81]

Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

5 0

3 years ago

Read 2 more answers

A capacitance C and an inductance L are operated at the same angular frequency.

Levart [38]

A) $\omega = \frac{1}{\sqrt{LC}}$

The magnitude of the capacitive reactance is given by

$X_C = \frac{1}{\omega C}$

where

$\omega$ is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

$X_L = \omega L$

where L is the inductance.

Since we want the two reactances to be equal, we have

$X_C = X_L$

So we find

$\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}$

B) 7449 rad/s

In this case, we have

$L=5.30 mH = 5.3\cdot 10^{-3}H$ is the inductance

$C= 3.40 \mu F= 3.40 \cdot 10^{-6}F$ is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

$\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s$

C) $39.5 \Omega$

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

$X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega$

- Inductive reactance:

$X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega$

7 0

3 years ago