Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a
1 answer:
Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
You might be interested in
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb =
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.
= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) =
=
=
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g =
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.