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3 years ago

The diagram above shows four light bulbs wired in a circuit. What would happen if bulb 3 burned out

Physics

1 answer:

andre [41]3 years ago

6 0

The lights will no longer work because the circuit will be cut

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A ball of mass 0.152 kg is dropped from a height 2.27 m above the ground. The acceleration of gravity is 9.8 m/s 2 . Neglecting

Natali5045456 [20]

Answer:

8.44 m/s.

Explanation:

Change in Potential Energy = Mass x Acceleration From Gravity x H2 - H1

Kinetic Energy = 1/2 (mass) x [(v2)^2 - v1^2]

g * h = 1/2 * v^2

(9.8) x (2.27) = 1/2 * (v)^2

v^2 = 2[(9.8) x (2.27)]

v = 6.67 m/s

g * delta h = 1/2 * delta v^2

(9.8) x (2.27 - 0.903) = 1/2 * [(v2)^2 - (6.67)^2]

v2^2 = 71.2821

v2 = 8.44 m/s

5 0

3 years ago

Read 2 more answers

If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out

Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0

2 years ago

The combined-gas law relates which of these?

Fofino [41]

The combined-gas law relates which temperature, pressure and volume.

Temperature=T
Pressure=P
Volume=V

(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.

5 0

3 years ago

A falcon can descend with a speed 250 km/h. If a falcon flies at this speed for 2.0 s and then flies a 100 m in 2.5 s, what is t

Vadim26 [7]

Answer:

v= s/t

Explanation:

250 km/ h =69.44m/s

S1=2 times 69.44 ≈ 139m

Next 2.5 seconds:

S2 = 100m

Average speed:

v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h

3 0

3 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago