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4 years ago

PLEASE HELP’!

Physics

1 answer:

salantis [7]4 years ago

3 0

Answer:

a= g sinθ

Explanation:

Given that

Angle ,θ = 15° (from the horizontal)

Lets take the mass of the block = m

The acceleration due to gravity = g

The acceleration of the block = a

When block slide down :

The gravitational force on the block = m g sinθ

By using Newton's second law

F= m a

F=Net force ,a acceleration ,m=mass

m g sinθ = ma

a= g sinθ

Therefore the acceleration of the block will be g sinθ.

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The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very

BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

$F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.

6 0

3 years ago

A hot-air balloon is descending at a rate of 2.1 m/s when a passenger drops a camera. If the camera is 42 m above the ground whe

zysi [14]

Answer:

a) 2.7s

b) 29 m/s

Explanation:

The equation for the velocity and position of a free fall are the following

$v=v_{0}-gt$ -(1)

$x=x_{0}+v_{0}t-gt^{2}/2$ - (2)

Since the hot-air ballon is <em>descending </em>at 2.1m/s and the camera is dropped at 42 m above the ground:

$v_{0}=-2.1m/s$

$x_{0}=42m$

To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:

$t = \frac{1}{84}(2.1\pm\sqrt{2.1^{2} - 4\times42\times9.81/2} )$

t = 2.71996

Rounding to two significant figures:

t = 2.7 s

Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s

$v=-2.1-9.81*(2.71996)$

v = -28.782 m/s

Rounding to two significant figures:

v = -29 m/s

where the minus sign indicates the downwards direction

3 0

3 years ago

A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel befor

Alja [10]

Answer:

Δx = 39.1 m

Explanation:

Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)$

where vf is the final velocity (0 in our case), v₀ is the initial velocity

(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

traveled since the brakes are applied.

Solving (1) for Δx, we have:

$\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)$

7 0

3 years ago

The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl

aleksandr82 [10.1K]

$acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}$

6 0

4 years ago

Whats a peer review?

Goryan [66]

B is the answer to your question

8 0

3 years ago

Read 2 more answers