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Alla [95]
3 years ago
12

How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?

Physics
1 answer:
Anna [14]3 years ago
7 0
Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.
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Individuals have different traits because they are not twins so they should not  be alike also  because people have different genes witch would give them different traits. For example not every individual has blonde hair and blue eyes because they did not inherit those traits.
3 0
3 years ago
The older, geocentric model of the solar system placed the__________ at the center. Today's heliocentric model places the_______
DedPeter [7]

the answer is earth for the first one and sun for the second one

3 0
3 years ago
Read 2 more answers
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
vitfil [10]

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: F_f=\mu mg, backward, with

\mu coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

F-F_f = 0

which we can solve to find the coefficient of kinetic friction:

F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20

7 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
3 years ago
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
Tanzania [10]

A) 18.4 m

B)

a) mass of the load

b) mass of the truck

Explanation:

A)

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g (1)

where

m is the mass of the load

\mu=0.400 is the coefficient of static friction

g=9.8 m/s^2 is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

v^2-u^2=2as (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m

B)

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck

8 0
3 years ago
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