Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me

An object at rest cannot remain at rest unless which of the following holds?View Available Hint(s)An object at rest cannot remai

Leni [432]

Answer:

True The net force must be zero for the acceleration to be zero

Explanation:

In order to analyze the statements of this problem we propose your solution.

First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)

Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.

As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.

Now we can analyze the statements given.

True The net force must be zero for the acceleration to be zero

False. If the force is different from zero, there is acceleration that changes the speeds

False. There may be forces, but the sum of them must be zero

False. If a force acts, the acceleration is different from zero and the speed changes

5 0

3 years ago

A proton has been accelerated from rest through a potential difference of -1000 v . part a what is the proton's kinetic energy,

wlad13 [49]

We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
$\Delta U + \Delta K=0$
which means
$\Delta K = - \Delta U$
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference ( $\Delta V=-1000 V$ ):
$\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV$
Therefore, the kinetic energy gained by the proton is
$\Delta K = -(-1000 eV)=1000 eV$
And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.

4 0

3 years ago

Read 2 more answers

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

7 0

3 years ago

Rosa uses the formula (vi¡cosθ)tΔ to do a calculation. Which value is Rosa most likely trying to find?

andriy [413]

To break this problem down, let's start with what we know. The equation given finds one component of the velocity and multiplies it by the change in time. This will not find the acceleration that the first two answers say it will, meaning that the answer isn't A or B.

That leaves us with the final two answers, C and D. If the projectile was launched horizontally and we were trying to find the horizontal displacement, we wouldn't need to use cosθ to find the horizontal velocity, meaning that our answer is most likely C) the horizontal displacement of a projectile launched at an angle!

4 0

3 years ago

A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to

maria [59]

Answer:

the distance between interference fringes increases

Explanation:

For double-slit interference, the distance of the m-order maximum from the centre of the distant screen is

$y=\frac{m \lambda D}{d}$

where $\lambda$ is the wavelength, D is the distance of the screen, and d the distance between the slits. The distance between two consecutive fringes (m and m+1) will be therefore

$\Delta y = \frac{(m+1) \lambda D}{d}-\frac{m \lambda D}{d}=\frac{\lambda D}{d}$

and we see that it inversely proportional to the distance between the slits, d. Therefore, when the separation between the slits decreases, the distance between the interference fringes increases.

4 0

3 years ago

Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me​

Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me