A ball connected to a 1.1 m string and is swing in circular fashion. It’s tangential velocity is 15 m/s. What is its centripetal acceleration?

Answers

Answer 1

Answer:

Answer:

ac = 204 [m/s²]

Explanation:

To solve this problem we must use the following equation that relates the tangential velocity to the radius of rotation.

ac = v²/r

where:

v = tangential velocity = 15 [m/s]

r = radius = 1.1 [m]

Now replacing we have:

ac = (15)²/1.1

ac = 204 [m/s²]

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A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.a. Determine the value of the frictional force exerted on the car.

b. Determine the value of the coefficient of friction between the tires and the road.

Answers

(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude a of

a = v ² / R

where

v = tangential speed

R = radius of the curve

so that

a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

∑ F = Fs = ma

where

Fs = magnitude of static friction

m = mass of the car

Then

Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N

(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

N - W = 0

where

N = magnitude of normal force

W = weight

so that

N = W = m g = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:

Fs = µN

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

Jolene travels north 5 miles and then goes west 3 miles before coming straight back south 2 miles. What is her distance

Answers

Answer:

mnbhngbfcvdxc

Explanation:

How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

$F_(cp)=f_a$

$(m.v^(2))/(R)=\mu.F_N$

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm What is the distance between the two second-order minima?

Answers

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=(y_1)/(2)\n\Rightarrow \beta_1=(1.4)/(2)\n\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=(m_1\lambda D)/(d)\n\Rightarrow d=(m_1\lambda D)/(\beta_1)$

For second order

$\beta_2=(m_2\lambda D)/(d)\n\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\n\Rightarrow \beta_2=(m_2)/(m_1)\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\n\Rightarrow y_2=2(m_2)/(m_1)\beta_1\n\Rightarrow y_2=2(2)/(1)* 0.7\n\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

Which two processes allow water to enter the atmosphere

Answers

The two processes that allow water to enter the atmosphere are: Evaporation and Transpiration.

1. **Evaporation:** Evaporation is the process by which liquid water on the Earth's surface (such as oceans, lakes, rivers, and even moist soil) is heated by the sun and turns into water vapor. This water vapor then rises into the atmosphere. Evaporation is a key component of the water cycle, where water constantly moves between the surface of the Earth and the atmosphere.

2. **Transpiration:** Transpiration is the process by which water is released into the atmosphere from plants. Plants take up water from the soil through their roots, and this water travels through the plant and eventually evaporates from small openings called stomata on the leaves. Transpiration serves various functions in plants, including cooling the plant and transporting nutrients.

Together, evaporation and transpiration contribute to the overall movement of water from the Earth's surface into the atmosphere, where it eventually condenses to form clouds and participates in various atmospheric processes before returning to the surface as precipitation through processes like rain, snow, sleet, or hail.

To know more about atmosphere:

brainly.com/question/32358340

#SPJ6

Answer:

sublimation and transpiration

Explanation:

The water can enter the atmosphere from snow and ice with the process of sublimation where they also make water vapors. Last water can get in the atmosphere from plants through transpiration which means that the water is evaporated through the pores of the leaves

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