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Tresset [83]
4 years ago
15

A stack of bricks weighs 170 KN (Kilo-newtons). The stack exerts 180 KPa (Kilo-pascals) of pressure on the ground. What is the a

rea upon which this pressure is exerted (in square ft)?
Physics
1 answer:
Alex17521 [72]4 years ago
5 0

Answer:

10.12square feet

Explanation:

Pressure exerted on the object is defined as the ratio of the force exerted on it to its unit area. Mathematically, Pressure = Force/Area

Given the force = 170kN

Pressure = 180KPa

Area = Force /Pressure

Area = 170kN/180KPa

Area = 0.94N/Pa

Note that 1Newton/Pascal = 10.764square feet

Therefore 0.94N/Pa = x

x = 0.94× 10.764

x = 10.12square feet

Therefore the area upon which this pressure is exerted is 10.12sqft.

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Two lamps rated 60W; 240V and 100W, 240Vrespectively are connected in series to a 240V power source. Calculate;
nikdorinn [45]

Answer:

See the answers below.

Explanation:

The total power of the circuit is equal to the sum of the powers of each lamp.

P=60+100\\P=160 [W]

Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.

P=V*I

where:

P = power [W]

V = voltage [V]

I = current [amp]

I = P/V\\I=160/240\\I=0.67 [amp]

So this is the answer for c) I = 0.67 [amp]

We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.

a)

V=I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohms]

Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.

P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R

60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm]

b)

The effective resistance of a series circuit is equal to the sum of the resistors connected in series.

R = 133.66 + 225\\R = 358.67 [ohms]

7 0
3 years ago
One simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching to
yaroslaw [1]

Complete Question:

One simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.5 m/s in 2.24 s, what will be his total time?

Answer:

total time = 6.24 s

Explanation:

Using the equation of motion:

v = u + at

initial speed, u = 0 m/s

v = 11.5 m/s

t = 2.24 s

11.5 = 0 + 2.24a

a = 11.5/2.24

a = 5.13 m/s²

For the total time spent by the sprinter:

s = ut + 0.5at²

100 = 0.5 * 5.13 * t²

t² = 100/2.567

t² = 38.957

t = √38.957

t = 6.24 s

3 0
4 years ago
a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e
natima [27]

Answer:

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J

5 0
3 years ago
Two identical pebbles are dropped. The first is dropped from a height of 256 feet and the second is dropped from a height of 400
ehidna [41]

Answer:

4.022 seconds and 4.99 seconds

Explanation:

Hello!

The free fall of the stone corresponds to a uniformly varied rectilinear movement

d=V_0*t+1/2*g*t^2

Being a free fall the initial speed is zero.

The distance is positive when considered in the same direction and direction as acceleration and speed.

256 feet stone

79.25 m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t = 4.022 seconds

400 feet stone

121.92m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t= 4,99 seconds

success with your homework!

Download pdf
8 0
3 years ago
Which example describes constant acceleration due only to a change in direction
Degger [83]
A car driving at a constant speed around a circular track
8 0
4 years ago
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