PHYSICS COLLEGE

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration.

Answers

Answer 1

Answer:

Answer:

(a) The average velocity is 16 m/s

(b) The acceleration is 0.4 m/s^2

(c) The final velocity is 24 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+(a.t^2)/(2)\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=(x)/(t)=(640)/(40)=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=(v_f-v_o)/(t)$

$\displaystyle a= 2(x-v_ot)/(t^2)$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2(640-8\cdot 40)/(40^2)=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

Answer 2

Answer:

Final answer:

The average velocity is 16 m/s, the final velocity is 8.0 m/s + (acceleration * 40 s), and the acceleration can be found by solving the equation 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2.

Explanation:

To find the average velocity, we use the formula: average velocity = total displacement / total time. In this case, the total displacement is 640 m and the total time is 40 s, so the average velocity is 640 m / 40 s = 16 m/s.

To find the final velocity, we can use the formula: final velocity = initial velocity + (acceleration * time). In this case, the initial velocity is 8.0 m/s and the time is 40 s. Since the question states that it moves with constant acceleration, we can assume that the acceleration is the same throughout the 40 s interval. Therefore, the final velocity is 8.0 m/s + (acceleration * 40 s).

To find the acceleration, we can use the formula: total displacement = (initial velocity * time) + (0.5 * acceleration * time^2). In this case, the total displacement is 640 m, the initial velocity is 8.0 m/s, and the time is 40 s. Solving for acceleration, we have 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2).

Learn more about Average velocity, Final velocity, Acceleration here:

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11.
A current of 67 amps runs through a resistor of 37 ohms, how much voltage is lost?

Answers

You divide them. 67/37 is 1.5 and so you subtract it with the 67 and multiple it by the coherent integer from the multiplication and you would get 20 volts lost roughly.

Answer: 20 volts

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

$1.6* 10^(-7)$ N

$2.4* 10^(-7)$ N

Explanation:

$i_(1)$ = 1 A

$i_(2)$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)$

$F = (10^(-7))(2(1)(4))/(5)$

$F = 1.6* 10^(-7)$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))$

$B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))$

$B = 2.4* 10^(-7)$

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Answers

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = (kq)/(r^2)$

k is the coulomb constant

$q= (Er^2)/(k)$

$q= (1.18* 0.822^2)/(9* 10^9)$

q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

Answers

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?

Answers

Answer:

r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

Em₀ = U_g + U_e + K = $-G (m_1m_2)/(r) +k (q_1q_2)/(r) - 2 ( (1)/(2) m v^2)$

the two in the kinetic energy is because they are two particles

final point. When it is detained

Em_f = U_g + U_e = $-G (m_1m_2)/(r_1) + k (q_1q_2)/(r_1)$

the energy is conserved

Em₀ = em_f

the charges and masses of the two particles are equal

$-G (m^2)/(r) + k (q^2)/(r) + m v^2 = - G (m^2)/(r_1) + k (q^2)/(r_1)$

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁

-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

r₁ = 2.045 10⁻¹ m

r₁ = 20.5 cm

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