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2 years ago

6

A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. If friction is negligible, what is the resu

ltant force on the sled?

1 answer:

slega [8]2 years ago

8 0

Answer:

Resultant force = 17.02 N

Explanation:

As we assume the coefficient of friction is negligible, the normal force won't affect the resultant force.

As a result of this, the sine of the angle times the force of gravity on the 10 kg mass is equal to the resultant force, which is, force due to gravity = m × a = 10 kg × 9.8 m/s² = 98 N

98 sin(10)=?

Sin(10)= 0.1736

98 x 0.1736= 17.02 N.

Therefore, the resultant force = 17.02 N

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A searchlight is 210 ft from a straight wall. As the beam moves along the wall, the angle between the beam and the perpendicula

Ivenika [448]

Answer:

The length of the beam increasing is 9.64 ft/s.

Explanation:

Given that,

Height = 210 ft

Distance =290 ft

According to figure,

We need to calculate the angle

$\cos\theta=\dfrac{210}{x}$ ....(I)

Put the value of x in the equation

$\cos\theta=\dfrac{210}{290}$

$\cos\theta=\dfrac{21}{29}=0.72$

Now, $\sin\theta=\dfrac{20}{29}$

On differentiate of equation (I)

$-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}$

$\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}$

Put the value in the equation

$\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}$

$\dfrac{dx}{dt}=9.64\ ft/s$

Hence, The length of the beam increasing is 9.64 ft/s.

4 0

2 years ago

An astronaut is rotated in a horizontal centrifuge at a radius of 5.0 m. (a) What is the astronaut’s speed if the centripetal ac

sergeinik [125]

Explanation:

Given that,

Radius of circular path, r = 5 m

Centripetal acceleration, $a=7g\ m/s^2$

(a) Let v is the astronaut’s speed. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$v=\sqrt{ar}$

$v=\sqrt{7\times 9.8\times 5}$

v = 18.5 m/s

(b) Let T denotes the time period. It is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi \times 5}{18.5}$

T = 1.69 s

Let N is the number of revolutions. So,

$N=\dfrac{60}{1.69}=35.5\ rev/min$

So, the number of revolutions per minute is 35.5

(c) T = 1.69 seconds

Hence, this is the required solution.

8 0

2 years ago

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

3 years ago

Read 2 more answers

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

#LearnwithBrainly

7 0

2 years ago

The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th

Bezzdna [24]

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

7 0

2 years ago