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yan [13]
3 years ago
10

A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initial

Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a =  -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

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Water pollution can affect the economy because with no clean water, people will be forced to buy water bottles impacting their disposable income. In the search for clean water, potential future conflicts may occur because water is a fundamental need to human survival. :) 
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3 years ago
A fan blade is rotating with a constant angular acceleration of +14.1 rad/s2. At what point on the blade, as measured from the a
slavikrds [6]

Answer:

0.695 m

Explanation:

α = angular acceleration of the fan blade = 14.1 rad/s²

a = tangential acceleration at the point concerned = acceleration due to gravity = 9.8 m/s²

r = distance of the point from axis of rotation at which tangential acceleration is same as acceleration due to gravity

We know the relation between

a = r α

Inserting the values

9.8 = 14.1 r

r = 0.695 m

6 0
3 years ago
The measure of an object's speed and direction is the object's ______.
V125BC [204]

Answer:

47 more rows

Explanation:

A B

speed rate of change of postion

velocity speed + direction of a moving object

weight measure of the force of gravity on an object

scale interment to measure weight

3 0
3 years ago
Read 2 more answers
3.00 m3 of a fixed mass of a gas at 150 kPa. Calculate the pressure if the volume is reduced to 1.20 m3 at a constant temperatur
viktelen [127]

Answer:

P₂ = 375 kPa.

Explanation:

Given that,

Initial volume, V₁ = 3 m³

Initial pressure, P₁ = 150 kPa

Final volume, V₂ = 1.2 m³

We need to find the final pressure.

At constant temperature,

P\propto \dfrac{1}{V}\\\\P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{150\ kPa\times 3\ m^3}{1.2\ m^3}\\\\P_2=375\ kPa

So, the new pressure is 375 kPa.

6 0
3 years ago
A plane flying horizontally at an altitude of 1 mi and a speed of 560 mi/h passes directly over a radar station. Find the rate a
natulia [17]

Given:

altitude, x = 1 mile

speed, v = 560 mi/h

distance from the station, x = 4 mi

Solution:

To find the rate,

\frac{dx}{dt} = 0

Now, from the right angle triangle in fig 1.

Applying pythagoras theorem:

h^{2}=x^{2} + y^{2}

differentiating the above eqn w.r.t 't' :

2h\frac{dh}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}                  (1)

Now, putting values in eqn (1):

2h\frac{dh}{dt} = 2\times 1\times 0 + 2y\frac{dy}{dt}

\frac{dh}{dt} = \frac{y}{h}\frac{dy}{dt}

\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}

\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}

\frac{dh}{dt} = 140\sqrt{4^2 - 1}

The rate at which distance from plane to station is increasing is:

\frac{dh}{dt} = 542.22 mph

6 0
3 years ago
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