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Marrrta [24]
3 years ago
6

An object of mass m slides down an incline with angle 0.

Physics
2 answers:
juin [17]3 years ago
8 0

Answer:

B). F_N - mgcos\theta = 0

Explanation:

As we know that the force on the mass "m" which slides along the inclined plane is along the inclined surface

So it must have accelerated along that surface only

So we can write the equation for force along the inclined plane as

F_{net} = ma

now if we wish to find net force perpendicular to the plane then we can say that it must be zero

Because perpendicular to the plane the net force on the given mass must be zero as it is not accelerating along that direction

So here we can say

F_{net} = 0

so we have

F_N - mgcos\theta = 0

MAVERICK [17]3 years ago
5 0
<h3>Answer</h3>

B)

FN - mgcosФ = 0

<h3>Explanation</h3>

Given the mass of an object, forces that will be acting on it are as following

  • Force due to gravity (weight) of an object
  • The Normal force (the perpendicular force from that the inclined surface exerts on the object)
  • Any frictional forces (due to static friction or kinetic friction, if applicable) along the incline
  • The corresponding force component along the incline (downhill force due to gravity)

Force due to gravity = -mgcosΔ

Normal force = fN

Frictional force = -f

force applied forward = mgsinΔ

Horizontal force = mgsinΔ - f

vertical force = fN -  mgcosΔ

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
If a town was 90 miles away and you travel at 45 mph how long would it take to get there
notka56 [123]

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

v=\dfrac{d}{t}

Where

t is time

t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h

So, the required time is 2 hours.

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3 years ago
How to convert 250 newton to pounds? It is likely to be difficult to move?
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3 years ago
Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi
max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact  momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

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3 years ago
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