A man is riding his 4-wheeler at 60 km/hr. If he is riding it constantly at this rate for 1.5 hours, how far did he ride?

1. Susan pushes her dad, David, on an ice rink with a force of 30 N. She weighs 45 kg and her dad weighs 100 kg. What are the ac

mamaluj [8]

Explanation:

Given parameters:

Force = 30N

Weight Susan = 45kg

Weight of Dad = 100kg

Unknown:

Acceleration of Susan = ?

Acceleration of Dad = ?

Solution:

Force = mass x acceleration

Acceleration = $\frac{force}{mass}$

Acceleration of Susan = $\frac{30}{45}$ = 0.67m/s²

Acceleration of Dad = $\frac{30}{100}$ = 0.3m/s²

Learn more:

Acceleration brainly.com/question/3820012

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8 0

3 years ago

PLEASE HURRY I HAVE A TIME LIMIT

RideAnS [48]

Properties of electromagnetic radiation and photons. ... we find the types of energy that are lower in frequency ( and thus longer in wavelength) than visible light. Seeee

4 0

3 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

Alec says the force of gravity is stronger on a piece of paper after it’s crumpled. His classmate, Jordan, disagrees. Alec “prov

Anika [276]

No, gravity acts equally on all objects. The crumpled paper falls faster because it resists the drag force due to the atmosphere because of its compact size. A flat piece of paper has an extended body and "catches" the air and falls more slowly. In a vacuum they would fall at the same rate either way.

8 0

3 years ago

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth d and the other (lets call it l) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section A at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed v of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where q is the water flow (1 cubic foot per minute).

7 0

3 years ago