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Hoochie [10]
3 years ago
12

the line on the position time graph show the velocites of different vehicles which line represents a vehicle moving at constant

velocity
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

For some mysterious reason, we can't see the graph.

On a position/time graph, constant velocity is represented by a <em>straight line</em>.  Depending on the velocity, the line may be sloping up, sloping down, or horizontal.  The only thing the line on the graph <u><em>CAN't</em></u> be is vertical.

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
Object A has a mass of 100 grams. Object B has a mass of 150 grams. They are both traveling at the same velocity. What can you c
Scilla [17]
-- Momentum is (mass) x (speed). 
Object B has 1.5 times as much momentum as Object A has.

-- Kinetic energy is (1/2) x (mass) x (speed) . 
Object B has 1.5 times as much kinetic energy as Object A has.

-- If they would both stop long enough to get on the scale,
Object B would weigh 1.5 times as much as Object A does.
8 0
3 years ago
330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°
Olenka [21]
The final temperature of the system is 32.5°
we know,  H = mcT 
where, H = Heat content of the body 
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry 

The net heat remains constant i.e. 
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x \frac{900}{4200} x (T - 10) 

⇒ 14,850 - 330T = 183.21T  - 1832 

⇒ - 513.21 T = - 16682

or T = 32.5°
3 0
3 years ago
when the mass of an object is increased it would decrease its acceleration in the force is left alone
mojhsa [17]

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

a=\frac{F}{m}

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

4 0
3 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0
3 years ago
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