y = 75.9 m
Explanation:
y = -(1/2)gt^2 + v0yt + y0
If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.
y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)
= -44.1 m + 120 m
= 75.9
Sound waves move through molecules. When there are no molecules in a certain space, sound cannot travel - whether in a vacuum or in space.
We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)
The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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Answer:
d. )directed upward.
Explanation:
As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.
This is because the electric field has the same direction that the force on a positive test charge at the same point.
As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.
So, the statement d. is the one that results to be true.
