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Dennis_Churaev [7]
3 years ago
11

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is dire

cted
a)vertically.
b)perpendicular to the plane.
c)at an angle θ ¸ below the inclined plane.
d)parallel to the plane in the same direction as the movement of the box.
e)parallel to the plane in the opposite direction as the movement of the box.
Physics
1 answer:
anzhelika [568]3 years ago
4 0

Answer:

vertically

Explanation:

The gravitational force on any object will always be vertically. (at an angle 90-θ below the inclined plane).

<em>The force made by the inclinated plane on the box and the force made by the box on the plane will be perpendicular to the plane.</em>

<em>The force made by the inclinated plane on the box and the force made by the box on the plane will be perpendicular to the plane.The acceleration of the box will be parallel to the plane in the same direction as the movement of the box, and the fricction force will be eparallel to the plane in the opposite direction as the movement of the box.</em>

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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0
3 years ago
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
Because weight is a measurement of the amount of gravity pulling on an object, weight is considered a ______. Question 18 option
Andrews [41]

Answer: Force I believe.

6 0
2 years ago
the volume of ice block is 2400cm^3 and its density is 0.9 g/cm^3. how much part of it remains above the surface of water when i
stiks02 [169]

Answer:

240 cm³

Explanation:

Weight = Buoyancy

mg = ρVg

m = ρV

(0.9 g/cm³ × 2400 cm³) = (1 g/cm³) V

V = 2160 cm³

The submerged volume is 2160 cm³, so the volume above the surface is 240 cm³.

4 0
3 years ago
H) Find the inverse Laplace transforms of the equation. When the X(s) =1 /(s²+9s+20)​
Anastasy [175]

Answer:

e^-4t e^-5t

Explanation:

solving s²+9s+20 quadratically we have (s+4)(s+5)

x(s) can be written as x(s) =(1/s+4)(1/s+5)

if we take the laplace inversve

L-¹ (s)=L-1(1/s+4) L-1(1/s+5)

we have e^-4t e^-5t

7 0
1 year ago
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