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Nataliya [291]
3 years ago
7

A block of mass M is placed on an inclined surface. The incline makes an angle theta with respect to the horizontal. What are th

e magnitude and direction of the normal force exerted on the block by the incline? Group of answer choices magnitude =LaTeX: Mgcos\theta M g c o s θ , direction: perpendicular to the incline surface magnitude = LaTeX: Mgsin\theta M g s i n θ , direction: perpendicular to the incline surface magnitude =LaTeX: Mgcos\theta M g c o s θ , direction: perpendicular to the surface of the earth magnitude = Mg, direction: perpendicular to the incline surface magnitude = Mg, direction: perpendicular to the surface of the earth
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

N = Mg cos\theta

Explanation:

As we know that the block is placed on the inclined plane

So here we know that the components of weight of the block are

F_x = Mgsin\theta along the inclined plane

above component of the weight will be counter balanced by friction force as it will be at rest

F_y = Mgcos\theta perpendicular to the inclined plane

above component is perpendicular to inclined plane surface and it will be counter balanced by the normal force

So here correct answer for normal force will be

N = Mg cos\theta

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
3 years ago
The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

image distance   v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f  = .02215

f = 45.15 cm .

4 0
3 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
What is one Pascal pressure? What is its unit?​
BabaBlast [244]

Explanation:

the force acting perpendicularly on unit area of surface

  1. unit=pascle .
3 0
2 years ago
Read 2 more answers
You walk 20 m north then 30 m west for a total timer four minutes what is the magnitude of your average velocity in (m/s)
Shalnov [3]

Answer:

The average velocity is 0.15 m/s

Explanation:

Use the definition of average velocity as the distance traveled divided the time it took.

Since the movement was on the plane from the origin (0, 0) to the point (-30, 20) corresponding to 30 m west and 20 m north, we calculate the distance using the distance between two points on the plane:

distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(-30)^2+20^2} =\sqrt{1300} \approx 36.06\,\,m

Then the magnitude of the average velocity can be estimated via the quotient between distance divided time, but since the units required are meters per second, we first convert the four minute time into seconds: 4 * 60 = 240 seconds.

Then the average velocity becomes:

v_{ave}=\frac{distance}{time} =\frac{36.06}{240} =0.15\,\,m/s

8 0
3 years ago
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