If several forces are exerted on a system, calculate the work done by each force, then multiply the results.

A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000

Zigmanuir [339]

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). 

We want a ratio of 3000000 to 1, or 3 million to 1.

Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
v= 99.999999999999% of the speed of light ~ speed of light
v = 3 x 10^8 m/s

8 0

4 years ago

How do you calculate the resultant velocity of the object in this image?

alexgriva [62]

Answer:

the resultant velocity is Zero

Explanation:

by the rule of adding and subtracting factors, we know that; when the force acting on an object is from east & north we add and with forces acting from South & west we minus.

Therefore:

1) List the forces down:

the 100m/s acting west is (-) while

the other 100m/s is acting in easterly direction

so it is (+)

2) Add the forces:

-100+100=0

therefore the answer is 0m/s for the the resultant velocity

hope I'm right

4 0

2 years ago

tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one

Llana [10]

We don't know how many of ANY color are in the bag right now, so there's no way to calculate an answer.

What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.

4 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers

Can someone please explain how to solve 13?

frozen [14]

Hello!

Everything you've done so far seems to be correct. However, you need to get rid of the sin. To do this, use s $sin^{-1}$ .

This is known as the inverse of sin, which is where you go from a fraction to the actual angle itself.

When you run $sin^{-1}(1/5)$ through a calculator, you get about 11.537 degrees, or θ = 11.537.

To verify this, all you need to do is run sin (11.537), and the calculator returns 0.2, which is what we're looking for.

Hope this helps!

8 0

3 years ago