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3 years ago

(iii) Suggest why the lamp is safe to use, even though its cable only has two wires.

Physics

1 answer:

Sergeu [11.5K]3 years ago

6 0

Answer:

it is safe because the make it so that the wires use a small amount of enerygy and electricity at the same time. this is safe because it will make it so that the eletrical current is not doverwelming the wires.

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PLEASE HELP ASAP!!!

kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0

4 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

4 years ago

Please i need detailed explanation

zubka84 [21]

Answer:

2Micro Farahds

Explanation:

Its in the picture.

I Hope it helps.

4 0

3 years ago

A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes agains

gavmur [86]

Answer:

Explanation:

According to work energy theorem

change in kinetic energy of truck = work done against it

work done against it = force x displacement

= - 850 x 8 = 6800 J

change in kinetic energy of truck = - 6800 J .

energy will be reduced by 6800 J

5 0

4 years ago

(iii) Suggest why the lamp is safe to use, even though its cable only has two wires.​

(iii) Suggest why the lamp is safe to use, even though its cable only has two wires.