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Luden [163]
1 year ago
15

Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows

at a constant acceleration. While coming to a stop, the driver looks in the rear-view mirror and notices that the box is not slipping. In what direction is the frictional force acting on the box?
Physics
1 answer:
Scorpion4ik [409]1 year ago
5 0

Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

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To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m
hoa [83]

Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

F= ?

W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

3 0
2 years ago
What is the magnitude of the relative angle φ
melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0
2 years ago
You place your hands over a steaming bowl of soup to warm them. Which type of heat transfer are you experiencing?
Pani-rosa [81]

There could be a little bit of conduction through the air that's between the soup and your hand.  But it's very small, because air is not a good conductor of heat.

It's mostly <em>convection</em> ... hot air and steam rising from the soup to your hand.

Then, of course, there HAS to be some conduction when the hot gases reach your hand ... their heat has to soak into your skin, and that's conduction.

8 0
2 years ago
Read 2 more answers
In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

g = \frac{GM}{R^2}

Where G means gravitational universal constant and M the mass of the planet

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now calculate the value of the time

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

The time taken for the rock to reach the surface is t = 87.58s

8 0
2 years ago
Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder
Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0
2 years ago
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