A waterwheel is an example of:

if C is the vector sum of A and B C=A+B what must be true about directions and magnitude of A and B if C=A+B? what must be true

Hunter-Best [27]

The vector sum is the algebraic sum if the two vectors have the same direction.

The sum vector is zero if the two vectors have the same magnitude and opposite direction

Vector addition is a process that can be performed graphically using the parallelogram method, see attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.

There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel

case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector

case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.

In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector

7 0

4 years ago

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What is the tension in the string if the turntable is held fixed, with the mass hanging from the pulley (i. e. , the mass is sta

Paladinen [302]

Weight of the mass is the tension in the string if the turntable is held fixed, with the mass hanging from the pulley

Weight is a force that acts at all times on all objects near Earth. The Earth pulls on all objects with a force of gravity downward toward the center of the Earth. Weight (symbolized w ) is a quantity representing the force exerted on a particle or object by an acceleration field, particularly the gravitational field of the Earth at the surface.

The tension force is defined as the force that is transmitted through a rope, string or wire when pulled by forces acting from opposite sides. The tension force is directed over the length of the wire and pulls energy equally on the bodies at the ends.

The direction of tension is the pull which is given the name tension. Thus, the tension will point away from the mass in the direction of the string/rope. In case of the hanging mass, the string pulls it upwards, so the string/rope exerts an upper force on the mass and the tension will be in the upper side.

Tension will act opposite the the weight of the mass attached in order to maintain equilibrium .

To learn more about weight here

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7 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, wha

Sonja [21]

The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, what will the pressure of a gas be after the temperature changes?

8 0

3 years ago

The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from t

olga nikolaevna [1]

Solution:

a) We know acceleration due to gravity, g = GM/r²

Differential change, dg/dr = -2GM/r³

Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²

My height, h=dr = 1.7 m

Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²

or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³

or, 5.9*100³*G²*M² = 2c⁶

or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms

Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.

b) This limit an upper limit ,we can tolerate smaller masses only.

4 0

3 years ago