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2 years ago

Which of the following would have an effect on an object's mass?

1 answer:

8 0

None of the above

mass is measurement of how much stuff in inside something.
if you freeze or heat an object, you merely change the state of the object. the mass is conserved.
if you change the elevation, nothing happens to the mass. the stuff will not leave the object.
so it is None of the above

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A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon

erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

$F = \frac{dP}{dt}$

$F = \frac{m(v_f - v_i)}{\Delta t}$

now we know that

$m = 60 kg$

$v_f = 3 m/s$

$v_i = 0$

$\Delta t= 0.2 s$

from above equation

$F = \frac{60(3 - 0)}{0.2} = 900 N$

so he will experience 900 N force in above case

5 0

2 years ago

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

sleet_krkn [62]

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

8 0

2 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

2 years ago

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

3 years ago

The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of

Nimfa-mama [501]

Answer:

a. $3.95\times10^{26}$ W

Explanation:

$T$ = temperature of the surface of sun = 5800 K

$r$ = Radius of the Sun = 7 x 10⁸ m

$A$ = Surface area of the Sun

Surface area of the sun is given as

$A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}$

$e$ = Emissivity = 1

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

$P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W$

4 0

3 years ago