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1 year ago

Unit of work is derived unit why

1 answer:

3 0

The unit for work J can also be written as Nm. Therefore, it is a derived unit as different SI units are needed to obtain its value. In this case, its derived from force and distance.

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for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

2 years ago

An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me

Tasya [4]

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

3 0

2 years ago

Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assu

Vera_Pavlovna [14]

Answer:

The partial pressure of H2 is 0.375 atm

The partial pressure of Ne is also 0.375 atm

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

Pressure of sealed container = 1.5 atm

Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

7 0

2 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

2 years ago

Unit of work is derived unit why​

Unit of work is derived unit why