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Lubov Fominskaja [6]

3 years ago

12

two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram

of forces and find the resultant ànd direction of the resultant.

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:gay

Explanation:

gay

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A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

3 years ago

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2m/s to 4 m/s?

Vitek1552 [10]

Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago

Read 2 more answers

For an object to slow down and stop due to friction while sliding across the floor which of the following would be true.

EastWind [94]

Answer:10-4

Explanation:

4 0

3 years ago

While on a sailboat at anchor, you notice that 15 waves pass its bow every minute. The waves have a speed of 6.0 m/s . Part A Wh

Alchen [17]

Answer:

Distance between two adjacent wave crests = 24m

Explanation:

Distance= speed × time

Distance traveled by waves in 60 seconds (15 crests)= 15 × distance

15 × distance = 6,0 (meters/second) × 60 seconds

distance = (360 meters) / 15 = 24 meters (between two adyacent waves)

8 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago