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Arte-miy333 [17]
3 years ago
14

The amount of force needed to sustain motion of a rock in outer space is

Physics
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

0 N

Explanation:

According to Newton's first law, object subjected to no force or 0 net force will stay in the same motion. In outer space, there's no gravity or air resistance. So if the net force is 0, a rock in motion would be still in motion forever without a need of an extra force to sustain it.

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A pesticide kills insects by disabling the ribosomes in their cells.
Stolb23 [73]

Answer:

cell division

7 0
3 years ago
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A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min
jekas [21]

Answer:

a)  1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans  = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

or

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

or

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}

or

P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))

on substituting the values in the above equation, we get

P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

or

P_1=118.93\ kPa

thus, gauge pressure at point 1 is 118.93 kPa

8 0
3 years ago
A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al
lukranit [14]

Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

3 0
3 years ago
Your favorite taco stand is 800 m away, and closes in 15 minutes. How fast do you have to run to make it on
Dimas [21]

You must run at least 53.3333333 meters a minute.

8 0
3 years ago
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A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. write a quadratic function equation
Andrei [34K]
During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
Therefore; using the equation;
S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
Therefore; the quadratic equation is 5t² - 16t + h =0
5 0
3 years ago
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