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Delvig [45]
3 years ago
7

Please hurry

Physics
1 answer:
viktelen [127]3 years ago
6 0

Answer:

The velocity will be "76.8 m/s".

Explanation:

The given values are:

Acceleration,

a = 2.4 m/s²

Time,

t = 32 seconds

By equation of motion,

⇒  v=u+at

On substituting the values, we get

⇒     =0+2.4\times 32

⇒     =0+76.8

⇒     =76.8 \ m/s

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Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generato
Volgvan

Answer:

The coil radius of other generator is 5.15 cm

Explanation:

Consider the equation for induced emf in a generator coil:

EMF = NBAω Sin(ωt)

where,

N = No. of turns in coil

B = magnetic field

A = Cross-sectional area of coil = π r²

ω = angular velocity

t = time

It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.

Therefore, EMF for both coils or generators will be:

EMF₁ = NBπr₁²ω Sin(ωt)

EMF₂ = NBπr₂²ω Sin(ωt)

dividing both the equations:

EMF₁/EMF₂ = (r₁/r₂)²

r₂ = r₁ √(EMF₂/EMF₁)

where,

EMF₁ = 1.8 V

EMF₂ = 3.9 V

r₁ = 3.5 cm

r₂ = ?

Therefore,

r₂ = (3.5 cm)√(3.9 V/1.8 V)

<u>r₂ = 5.15 cm</u>

3 0
3 years ago
What traction of the radioisotope<br>remains in the body after one day?​
r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is (\frac{1}{2})^{\frac{1}{\tau}}, with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}

where:

m_0 is the mass of the radioactive sample at t = 0

\tau is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}

where the half-life \tau must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0
3 years ago
A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert
dlinn [17]
First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction) 

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

4 0
3 years ago
Read 2 more answers
Rest and motion are relative terms why?​
krok68 [10]

Answer:

Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.

3 0
3 years ago
What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?
ra1l [238]
A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour

Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

8 0
3 years ago
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