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sattari [20]
3 years ago
10

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Physics
1 answer:
Alexandra [31]3 years ago
6 0
600/3 = 200
the slope is 200m/min
 
OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour 

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Metals are used in many products because of the characteristics properties that most metals have. Which product requires the hig
nadezda [96]
Alright sounds good buddy
5 0
2 years ago
A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the ma
jonny [76]

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

h = ut+\dfrac{gt^2}{2}

Putting all given values, we get :

12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3

t = 2.56  and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

7 0
3 years ago
Which changes in energy form are illustrated in the diagram.
Grace [21]
It would be B since it starts with the solar energy which is converted to electricity with the solar panels, which then creates mechanical energy for the fans blades to move and sound for the radio.

Hope that helps :)
6 0
3 years ago
What law states force is dependent on the mass and acceleration of an object
UNO [17]

Answer:

Newton's second law of motion

Explanation:

Newton's second law of motion can be stated  

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

in another form,

Force = mass * acceleration

5 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
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