Why do heavier objects roll down slopes faster than lighter objects? (Assuming they are the same shape)

Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t

Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0

3 years ago

A 21 kg chair initially at rest on a horizontal floor requires a 175 N horizontal force to set it in motion. Once the chair is i

Alika [10]

Answer: $\mu =0.85$

Explanation:

Given

mass of chair m=21 kg

Force required to set chair in motion is 175 N

Once the chair is in motion 137 N is require to move it with constant velocity

i.e. 175 N is the amount of force needed to just overcome static friction and 137 is the kinetic friction force

thus

$f_s(static\ friction)=\mu \cdot N$

where $\mu$ is the coefficient of static friction and N is Normal reaction

$N=mg$

$f_s=\mu mg$

$\mu mg=175$

$\mu =\frac{175}{21\times 9.8}$

$\mu =0.85$

8 0

3 years ago

An automobile tire having a temperature of −1.6 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 22 lb/in2 . What

r-ruslan [8.4K]

Answer:

25.8 lb/in²

Explanation:

Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} } \\\\\frac{22}{-1.6+273.15} =\frac{P_{2} }{45+273.15} \\\\P_{2} = \frac{22*318.15}{271.55} = 25.8lb/in^{2}$

4 0

3 years ago

Which of the following is always true about a synthesis reaction?

djverab [1.8K]

The answer would be A

4 0

3 years ago

Read 2 more answers

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip

Murrr4er [49]

Answer:

Explanation:

Given that the weight of the crate is

M=48kg

Then, weight of boy is

W=mg=48×9.8=470.4N

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

N=W

The normal equals weight of boy

N=470.4N

Then, applying frictional force opposing the boy motion.

Fr=μN

Fr=0.5×470.4

Fr=235.2N

Then, this is the forward force that the boy try using to pull the crate.

Since he can barely move his foot he has not yet overcome the coefficient of static friction.

ΣF = ma. , along x-axis

a along x-axis is 0 since he cannot move his foot, I.e he was not moving

F-Fr= 0

F=Fr=235.2N

The forward force the boy apply is 235.2N on the crate

Also, analysing the crate.

The forward force is now F=235.2N

Then the frictional force =Fr

Then,

ΣF = ma. , along x-axis

a along x-axis is 0 since the crate did not move,

F-Fr=0

Fr=F=235.2N

This is the frictional force on the crate.

Then using frictional law

Fr=μN

N=Fr/μ

N=235.2/0.9

N=261.33N

Now this normal is equal to the weight of the crate

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

W=N=261.33N

Then, since weight is given as

W=mg

m=W/g

m=261.33/9.81

m=26.64kg

The mass of the crate is 26.64kg

7 0

3 years ago

Read 2 more answers