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weeeeeb [17]
3 years ago
8

A 900 kg car pushes a 2400 kg truck that has a dead battery. when the driver steps on the accelerator, the drive wheels of the c

ar push against the ground with a force of 5000 n. (a) what is the magnitude of the force of the car on the truck? n (b) what is the magnitude of the force of the truck on the car? n
Physics
1 answer:
Wewaii [24]3 years ago
5 0
A)0
b)magnitude of the force on the car is 500n
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You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double
dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

\tan\theta=\dfrac{y}{D}

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

\tan\theta=\dfrac{1.52}{120}

\theta=\tan^{-1}\dfrac{1.52}{120}

\theta=0.725

We need to calculate the wavelength

Using formula of wavelength

d\sin\theta=n\lambda

Put the value into the formula

0.025\times\sin0.725=5\times\lambda

\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}

\lambda=6.326\times10^{-7}\ m

\lambda=633\ nm

Hence, The wavelength of the light is 633 nm.

4 0
3 years ago
What is a creative name for an element? (like oxygen, phosphorus) try to make it good
Inga [223]
Oxyles i think this suite Oxyles what do you think.
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3 years ago
An object traveling around another object in space is in
drek231 [11]
It is orbiting the object.
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4 years ago
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

d=v_{0}.t + \frac{1}{2} .a.t^{2}, where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + \frac{1}{2}.10.t^{2}

h = 5t²

t_{T} = \sqrt{\frac{h}{5} }

On the other planet:

h =  0.t + \frac{1}{2}.30.t^{2}

h = 15.t²

t_{P} = \sqrt{\frac{h}{15} }

Comparing the 2 planets:

\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }

\frac{t_{T} }{t_{P} } = \sqrt{3}  or t_{T} = \sqrt{3}.t_{P}

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

5 0
3 years ago
Help ASAP thanks !!!
Ann [662]

Answer:

true

Explanation:

3 0
3 years ago
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