Question 2 (1 point)

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

2 years ago

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created is $P_{avg} = F * v_{avg}$

Explanation:

From the question we are told that

The that the average power is mathematically represented as

$P_{avg} = \frac{W }{\Delta t }$

Where W is is the Workdone which is mathematically represented as

$W = F * s$

Where F is the applies force and s is the displacement due to the force

So

$P_{avg} = \frac{F *s }{\Delta t }$

Now this displacement can be represented mathematically as

$s = v_{avg} * \Delta t$

Where $v_{avg }$ is the average velocity and $\Delta t$ is the time taken

So

$P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }$

=> $P_{avg} = F * v_{avg}$

3 0

2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago

A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?

Shtirlitz [24]

Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.

8 0

2 years ago

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

2 years ago