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rewona [7]
3 years ago
14

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

e velocity and speed (in m/s) when t = 5. f(t) = 18 + 48/t + 1
Physics
1 answer:
Rashid [163]3 years ago
6 0

Answer:

The velocity of the particle = -1.92 m/s

The speed of the particle = 5.72 m/s

Explanation:

Given equation of motion;

f(t) = 18 \ + \ \frac{48}{t} \ + \ 1

Velocity is defined as the change in displacement with time.

V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s

The distance traveled by the particle in 5 s:

s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m

The speed of the particle when t = 5s

Speed = \frac{28.6}{5} = 5.72 \ m/s

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How many electrons pass a given point in the circuit in 3 min? The fundamental charge is 1.602 × 10−19 C. amps: 0.415 , volts: 1
morpeh [17]

The no of electron passing through the circuit is current by time. The no of electron passing are  46.63 x 10¹⁹ electrons.

<h3>What is charge?</h3>

The charge is the physical quantity which attracts or repels another object when comes into its field.

The time 3 min has 3 x 60 =180 seconds.

Total Charge Q = current I x time t

Q = 0.415 x 180

Q =74.7 coulombs

The no of electrons is the ratio of total charge divided by the fundamental charge.

n = 74.7/  1.602 × 10⁻¹⁹

n= 46.63 x 10¹⁹ electrons

Thus, electrons pass a given point in the circuit in 3 min are 46.63 x 10¹⁹ .

Learn more about charge.

brainly.com/question/11944606

#SPJ1

4 0
2 years ago
A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?
Artist 52 [7]
The formula for momentum is mass times velocity. Simply, we just multiply the given values:
p = mv
p = 40 kg x 4 m/s
p = 160 kg m/s

Other units for momentum is N s.
p = 160 N s 
4 0
3 years ago
Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

4 0
3 years ago
Anyone please help me with this question ASAP
Katyanochek1 [597]

Answer:

increases

Explanation:

it would have to work harder to get to two points together

7 0
3 years ago
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