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3 years ago

11

Static___ sliding rolling_ fluid_____

2 answers:

Arisa [49]3 years ago

6 0

Answer:

a car

A sled sliding across snow or ice.

a ball down a hill

mercury

Explanation:

Phantasy [73]3 years ago

3 0

Static reaction

Sliding down a water slide

Rolling down a hill

Fluids running down a pipe

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A basket has a mass of 5.5 kg. Find the magnitude of the normal force if the basket is at rest on a ramp inclined above the hori

storchak [24]

Here is the correct answer of the given problem above.
Given that the basket has a mass of 5.5kg, the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal is at 12 degrees. The solution is simple:
<span>Fn at rest = lmgl </span>
<span>= 5.5kg (9.80N/kg)
=</span><span> mgCos12degrees
Hope this answer helps. </span>

8 0

4 years ago

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Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

3 years ago

The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t

Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of $8.1\times 10^5 N$

Mass of train $=1.9\times 10^7$

Final speed (v) $=68 km/h \approx 18.88 m/s$

F=ma

so $acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}$

$a=0.042631 m/s^2$

and acceleration is

$a=\frac{v-u}{t}$

$0.042631=\frac{18.88-0}{t}$

$t=443.089 \approx 443.1 s$

8 0

3 years ago

Read 2 more answers

200 garms into kilogram

lana66690 [7]

Answer:

200/1000=0.2kg hope ur help and mark me brainlist

6 0

3 years ago

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What is the mass of an object that has a weight of 2867 N?

Gnom [1K]

Answer:

F = M a

W = M g equivalent equation to express weight of object of mass M

M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg

7 0

3 years ago