What is the formula for work?

A ball is kicked from the ground into the air at a velocity of 3 m/s at an angle 35° above the horizontal and hits the ground so

Dennis_Churaev [7]

Answer:2.45 m/s

Explanation:

Given

Launch velocity $(u)=3 m/s$

launch angle $=35^{\circ}$

as the vertical velocity first decreasing to zero and then increases to original value so its avg is zero .

$v_{avg}=\frac{displacement}{time}$

$v_{avg}=\frac{Range}{time}$

$v_{avg}=\frac{u\cos \theta \times t}{t}$

thus $v_{avg}=\frac{3\cos 35\times t}{t}$

$v_{avg}=3\cos 35=2.45 m/s$

8 0

3 years ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

4 0

4 years ago

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

victus00 [196]

Electric field lines always begin at positive charges (or at infinity) and end at negative charges (or at infinity).

One could also say that the lines we use to represent an electric field indicate the direction in which a positive test charge would initially move when released from rest.

6 0

4 years ago

When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is incre

Crazy boy [7]

Answer:

the previous correct answer is b

Explanation:

When the circuit is closed in the system, a current is induced that follows the lenz law, which opposes the change that is occurring and therefore the coil increases and the idicidal current in the ring must reach the maximum oppositing is the current of the coil, so quiet force is repulsion

Consequently, the previous correct answer is b

3 0

3 years ago

In the following lever system, what is the clockwise torque? 18 Nm 240 Nm 360 Nm 480 Nm

MaRussiya [10]

Answer:

240 Nm

Explanation:

The clockwise torque is the torque determined only by the force that makes the lever rotating clockwise: therefore, the force of 80 N on the right.

The torque produced by this force is given by:

$\tau = F d$

where

F is the magnitude of the force

d is the arm

For the force of 80 N on the right,

F = 80 N

d = 3 m (distance from the pivot)

So, the clockwise moment is

$\tau = (80)(3)=240 Nm$

3 0

3 years ago

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