Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:
With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:
This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.
Dispersion angle = 0.3875 degrees.
Width at bottom of block = 0.09297 cm
Thickness of rainbow = 0.07038 cm
Snell's law provides the formula that describes the refraction of light. It is:
n1*sin(θ1) = n2*sin(θ2)
where
n1, n2 = indexes of refraction for the different mediums
θ1, θ2 = angle of incident rays as measured from the normal to the surface.
Solving for θ2, we get
n1*sin(θ1) = n2*sin(θ2)
n1*sin(θ1)/n2 = sin(θ2)
asin(n1*sin(θ1)/n2) = θ2
The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
Red:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.641) = θ2
asin(1.00029*0.653420604/1.641) = θ2
asin(0.398299876) = θ2
23.47193844 = θ2
Violet:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.667) = θ2
asin(1.00029*0.653420604/1.667) = θ2
asin(0.39208764) = θ2
23.08446098 = θ2
So the dispersion angle is:
23.47193844 - 23.08446098 = 0.38747746 degrees.
Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
tan(θ) = X/11.6
11.6*tan(θ) = X
So for Red:
11.6*tan(θ) = X
11.6*tan(23.47193844) = X
11.6*0.434230136 = X
5.037069579 = X
And violet:
11.6*tan(θ) = X
11.6*tan(23.08446098) = X
11.6*0.426215635 = X
4.944101361 = X
So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm
The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
cos(θ) = X/0.092968218
0.092968218*cos(θ) = X
0.092968218*cos(40.80) = X
0.092968218*0.756995056 = X
0.070376481 = X
So the distance between the red and violet rays is 0.07038 cm.
Answer:
Explanation:
Distance between plates d = 2 x 10⁻³m
Potential diff applied = 5 x 10³ V
Electric field = Potential diff applied / d
= 5 x 10³ / 2 x 10⁻³
= 2.5 x 10⁶ V/m
This is less than breakdown strength for air 3.0×10⁶ V/m
b ) Let the plates be at a separation of d .so
5 x 10³ / d = 3.0×10⁶ ( break down voltage )
d = 5 x 10³ / 3.0×10⁶
= 1.67 x 10⁻³ m
= 1.67 mm.
If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.
Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.
If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.
Learn more: brainly.com/question/11401513
