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tankabanditka [31]
3 years ago
14

Choose the true statement concerning the mall or subatomic particles:

Physics
2 answers:
s344n2d4d5 [400]3 years ago
6 0
Hi there :)


Your answer is C. Protons and neutrons have about the same mass


Hope this helps

-Ans-
solniwko [45]3 years ago
4 0
C is true, and just one of those has as much mass as about 1,840 electrons.
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<span>A substance that accelerates the rate of chemical reaction is called a catalyst. It serves as an alternative pathway for the reaction product. The increase of rate of chemical reaction is because catalyst has low activiation energy than the original pathway. The low activation energy will increase the amount of molecules that can benefit in the energy created by the catalyst.</span>
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A rock is being twirled in a circle on the end of a string. The string provides the centripetal force needed to keep the ball mo
KonstantinChe [14]

Answer:

No

Explanation:

The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.

However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.

As we know, the work done by a force on an object is

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the force and the displacement

In this situation, F and d are perpendicular, so \theta=90^{\circ}, therefore cos \theta = 0 and the work done is zero:

W=0

4 0
2 years ago
How can you find the reading of main scale and vernier scale​
anygoal [31]

Answer:

this pdf should help you out

Explanation:

Download pdf
7 0
2 years ago
What is the momentum of a bird with a mass of 0.018kg flying at 15m/s?
Mazyrski [523]

Answer:

0.27 kg-m/s

Explanation:

i believe this is the correct answer

6 0
2 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
3 years ago
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