Answer:
speed-time graph is the acceleration of the particle or straight lines acceleration is constant
a distance-time graph represents that a particle has constant speed.
Explanation:
Answer:
Animals must eat other plants or animals to obtain the <u>energy</u> stored in the food
Explanation:
One classification of living organisms, according to the way they obtain energy, is that of autotrophs and heterotrophs. The first group is represented by plants, which process their own nutrients from inorganic matter.
<u>Animals -heterotrophes- are unable to process their own nutrients</u>, so they must obtain them from other organisms, either plants or animals. External food sources provide them with nutrients, which contain the energy substrate needed to perform their vital functions.
Learn more:
Autotrophs and heterotrophs brainly.com/question/7695115
Answer:
288N
Explanation:
Given parameters:
Mass of Cheetah = 12kg
Acceleration = 24m/s²
Unknown:
Force needed by the cheetah to run = ?
Solution:
The force needed by the Cheetah to run is the net force.
According to Newton's law;
Force = mass x acceleration
Insert the given parameters and solve;
Force = 12 x 24 = 288N
Answer:
a) t = 1.47 h b) t = 1.32 h
Explanation:
a) In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed
v =
f -
v = 585 -32.1
v = 552.9 km / h
We use the speed ratio in uniform motion
v = x / t
t = x / v
t = 815 /552.9
t = 1.47 h
b) We repeat the calculation, but this time the wind is going in the direction of the plane
v=
f -
v 585 + 32.1
v = 617.1 km / h
t = 815 /617.1
t = 1.32 h
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;

Therefore, the frequency of this mode of vibration is 138.87 Hz