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Morgarella [4.7K]
3 years ago
9

Numerical

Physics
1 answer:
ira [324]3 years ago
5 0

Answer:

Thus, 4 kg mass must be suspended from a steel wire 2 m long and 1 mm diameter to stretch it by 1 mm.

Explanation:

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if the resistance of a car headlight is 15 ohm and the current through it is 0.60, what is the voltage across the headlight?
Strike441 [17]

Answer:

9 volts (assuming 0.60 is in Amperes)

Explanation:

Recall that Ohms law can be expressed as

V = IR, where

V = voltage,

I = current (given as 0.6. I'm going to assume that the units is Amperes because it is not given)

R = resistance (given as 15 ohm)

substituting the above values into the formula

V = IR

V = (0.6)(15)

V = 9 Volts

4 0
3 years ago
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A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required t=\frac{distance}{speed}=\frac{d}{38}hour

Speed when car return from hill = 66 km/h

So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

8 0
3 years ago
Plants make food through photosynthesis, a chemical reaction. What are the starting substances of the reaction?
vekshin1
The answer is the 4th option because they don't use oxygen, that's what they produce
3 0
3 years ago
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What is the density of a rock with a mass of .235kg
Andre45 [30]

Answer:

where is volume? formula of density is: mass/volume so volume must be there

4 0
2 years ago
At which point on this electric field will a test charge show the maximum strength?
Juli2301 [7.4K]

Answer:

A

Explanation:

The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:

E\propto \frac{1}{r^2}

More precisely, the strength of the field at a distance r from the centre of the sphere is

E=k\frac{Q}{r^2}

where k is the Coulomb's constant and Q is the charge on the sphere.

From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.

This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.

8 0
3 years ago
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