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Paraphin [41]
3 years ago
11

When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4

Physics
1 answer:
vitfil [10]3 years ago
8 0
Just divide the two numbers with each other.
I mean 13/1.4=9.2857...
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Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. w
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Answer:

Zoning laws establish b. the uses an area of land can be put to.

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A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o
Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

v(t)=v_0 -gt

where

v_0=0 is the initial vertical velocity (zero, since the motorcycle is not moving)

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t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s

And since we know its mass, m=100 kg, we can find its momentum:

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and the negative sign simply means downward direction.

8 0
3 years ago
With speeds up to 90 miles per hour, what is the fastest sport on ice?.
Eddi Din [679]

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8 0
3 years ago
When you set something down on the ground wht kind of work are your arms doing?
umka2103 [35]
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7 0
3 years ago
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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

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Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
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