Answer:
The vehicle travels 56.25 metres in the interval during which body decelerates .
Explanation:
- Initial velocity of vehicle, u = 32 m/s
- Final velocity of vehicle, v = 22 m/s
- Rate of acceleration, a = - 4.8 m/
Let the distance travelled be s .
We have to determine the distance travelled by the vehicle during this time.
The equation of motion is given by
s =
<u>s = 56.25 metres</u>
The vehicle travels 56.25 metres in the interval during which body decelerates .
The given statement "When the net force of an object decreases, the object's acceleration decreases" is true about Newton's second law.
Answer: Option B
<u>Explanation:</u>
Newton’s second law of motion states that the object’s acceleration depends on two variable:
- Directly proportionate to the object’s force existed
- Inversely proportionate to the mass of the objects
The equation can be given as below,
Force = mass × acceleration
Hence, from the above, it is concluded that force is directly proportionate to mass and acceleration of the objects. So, when force increases both mass and acceleration increases. Similarly, if force decreases, both mass and acceleration get decreased. Therefore, the given statement in option B would be correct answer.
Answer:
photosynthesis uses carbon dioxide
Explanation:
Consider the following statements:
1. Carbon dioxide, chlorophyll, and sunlight are all essential for photosynthesis.
2. Rate of photosynthesis is maximum in red light and minimum in green light.
3. Increase
concentration decreases the rate of photosynthesis.
Which of the statements given above are correct about photosynthesis?
Answer:
Recall the Diffraction grating formula for constructive interference of a light
y = nDλ/w Eqn 1
Where;
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10-9 m
and n = 1
y₁ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1) = 0.994m
Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10⁻⁹ m
and n = 2
y₂ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m
Increased food production due to improved agricultural practices, control of many diseases by modern medicine and the use of energy to make historically uninhabitable areas of Earth inhabitable are examples of things which can extend carrying capacity.