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3 years ago

Please help on this one ?

Physics

1 answer:

satela [25.4K]3 years ago

5 0

It i c. because Newton's law was all about how gravity pulls us downward so that was going to be a difficulty in the Apollo mission.

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The bands on Jupiter are ultimately caused by...

Margaret [11]

Precision because it’s the answer

7 0

3 years ago

All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outle

VARVARA [1.3K]

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage ......................1

here voltage is same in all three device

current by toaster is

I = $\frac{1140}{120}$

I = 9.5 A

and

current by blender

I = $\frac{270}{120}$

I = 2.25 A

and

current by lamp is

I = $\frac{80}{120}$

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

8 0

3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago

The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

vitfil [10]

The expression of the electric flux is

$\Phi = \frac{Q}{\epsilon_0}$

Here,

Q = Total charge enclosed in the closed surface

$\epsilon_0$ = Permittivity due to free space

Rearranging to find the charge,

$Q = \epsilon_0 \Phi$

Replacing with our values we have finally

$Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)$

$Q = 1.6284*10^{-8} C$ $(\frac{10^9nC}{1C})$

$Q = 0.1684nC$

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0

3 years ago

A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

3 years ago