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vova2212 [387]
3 years ago
13

Which diagram shows the most likely effect when a rock is weathered by water flowing over its entire surface

Physics
1 answer:
balu736 [363]3 years ago
4 0

Answer:

the answer to this would be D.

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Uploaded imageTwo long, parallel wires each carry the same current I in the same direction (see figure below). The total magneti
LekaFEV [45]

Answer:

Zero

Explanation:

Two long parallel wires each carry the same current I in the same direction. The magnetic field in wire 1 is given by :

B_1=\dfrac{\mu_oI_1}{2\pi r}

Magnetic force acting in wire 2 due to 1 is given by :

F_{2}=I_2lB_1

F_2=\dfrac{\mu_oI_1I_2 l}{2\pi r}

Similarly, force acting in wire 1 is given by :

F_1=\dfrac{\mu_oI_1I_2 l}{2\pi r}According to third law of motion, the force acting in wire 1 will be in opposite direction to wire 2 as :

F_2=-F_1

So, the total magnetic field at the point P midway between the wires is in what direction will be zero as the the direction of forces are in opposite direction.  

7 0
3 years ago
Which of the following is an example of rotation
SOVA2 [1]
I believe it is D. Earth spinning on it's axis.
8 0
3 years ago
In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th
Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.  

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0
3 years ago
Read 2 more answers
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
3 years ago
What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s
stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0
3 years ago
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