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yanalaym [24]
2 years ago
7

At the end of the school day, at exactly 2:30 pm, a group of students run out of the school building and reach the edge of the s

chool property at 2:30:45 s. Which of the following correctly describes the motion in terms of time?
Question 2 options:

A)

t1= 0, t2 = 2:30:45 s, ∆t = 45 s

B)

t1= 2:30, t2 = 45 s

C)

∆t = 2:30

D)

t2= 2:30:45 s, ∆t = 45 s
Physics
1 answer:
Fofino [41]2 years ago
3 0

Explanation:

Initial time, t₁ = 2:30 pm

Final time, t₂ = 2:30:45

We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.

Change in time,

\Delta t=t_2-t_1\\\\\Delta t=2:30:45-2.30\\\\\Delta t=45\ s

Hence, this is the required solution.

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An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
2 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
2 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
swat32

Point X lies on a horizontal line. We can intuitively say that the slope of the graph at point X is 0, therefore the acceleration at point X is 0m/s²

Point Y lies on a downward slanting line. To calculate the slope of that line, let's apply this equation:

m = (y₂-y₁)/(x₂-x₁)

m = slope, (x₁, y₁) and (x₂, y₂) correspond to the coordinates of the line's endpoints.

Given values:

(x₁, y₁) = (7, 5)

(x₂, y₂) = (12, 0)

Plug in and solve for m:

m = (0 - 5)/(12 - 7)

m = -1

The acceleration at point Y is -1m/s²

Choice A

5 0
2 years ago
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The answer is 2 m/s east
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2 years ago
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