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DENIUS [597]
3 years ago
12

What force would be required to accelerate a 1,100kg car to 0.5 m/s2

Physics
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

the force required to accelerate a 1,100kg car is 550N

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Question 5 (Multiple Choice Worth 3 points) (02.07 MC) Rachel needs to eat fewer carbohydrates to improve her health. Which of t
patriot [66]
Milk, apples, and beans don't have much carbohydrate.  So if you
cut down on those, you don't really cut down much on carbohydrates.

If Rachel needs to reduce her intake of carbohydrates, she should
cut down on bread.  (Also cake, sugar, corn, pasta, and potatoes.)
5 0
3 years ago
A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

8 0
4 years ago
Give 5 examples of physical properties that could be used to describe a solid
ycow [4]
1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor
8 0
3 years ago
I need help ASAP I need to get this right plz plz plz!!!!!
zhenek [66]

Answer:

option d and b..............

3 0
3 years ago
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0
2 years ago
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