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3 years ago

5. How many kilowatt-hours of energy would be used by a 40 W bulb that runs for 10 hours every

Physics

1 answer:

Nady [450]3 years ago

6 0

Answer:

E = 0.4 kWh

Explanation:

Given that,

The power of a bulb, P = 40 W

Time, t = 10 hours

We need to find the energy used by the bulb. Let it is equal to E. So,

$E=P\times t\\\\E=40\times 10\\\\E=400\ W-h\\\\or\\\\E=0.4\ kWh$

So, 0.4 kWh of energy would be needed.

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Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible

storchak [24]

Answer:

no its not like the undertow in the ocean

Explanation:

4 0

3 years ago

A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, $u=32$ m/s

Acceleration of the driver is, $a=-1.5$ m/s²

Time taken to reach final velocity is, $t=10.8$ s.

The final velocity is given using the Newton's equations of motion as:

$v=u+at$ , where, $v$ is the final velocity.

Now, plug in the given values and solve for $v$ .

$v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}$

Therefore, his final velocity is 15.8 m/s.

5 0

3 years ago

The mass of a lift is 600kg.the. The maximum tensile force that the cable supporting the lift can withstand is 7kN. Calculate th

Oksi-84 [34.3K]

Force is the product of mass and acceleration .
The question is ask to find acceleration.
But acceleration is the ratio of the force and the mass.
where 600kg is the mass and 7kN is the force
NB: kilo is 1000
now we have to multiply 7N by 1000
by doing so you will have 7000N
which is the force.
Now to find the acceleration: force/ mass
which is 7000/600
therefore the maximum acceleration is 11.667

5 0

3 years ago

Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis

Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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