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Setler79 [48]
3 years ago
7

You are trying to move a chair of mass 29.0 kg by pushing it 30* below horizontal, but it is not sliding very easily. The coeffi

cient of kinetic friction between the chair and the floor is 0.625 and the resistive force is 283.6N.
(a) What is the normal force?
(b) What is the value of the angled force that will keep the chair moving at a constant velocity?
Physics
1 answer:
valina [46]3 years ago
7 0
A 3,000-N force gives an object an acceleration of 15 m/s2. The mass of the object is ____.

200 kg

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If a 500-pound object is moved 200 feet how much work is being done?
Paha777 [63]

Answer:

D

Explanation:

Work = Distance x Mass

4 0
3 years ago
The histogram below shows the number of downloads of a song over time.
Allisa [31]

Given data:

  • It is a graphical display where the data is grouped in to ranges
  • A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
  • It is an accurate representation of the distribution of numerical data.

<em>From Figure:</em>  

        Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).

<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>

7 0
3 years ago
A skateboarder is skating back and forth on the halfpipe as seen below. As he skates his energy transforms from potential energy
egoroff_w [7]

Answer:

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

Explanation:

The law of conservation of energy states that in absence of frictional forces, the mechanical energy of an object (given by the sum of its kinetic and potential energy) is conserved. In such a situation, the skateboarder would never stop his motion, because potential energy is continuously converted into kinetic energy and vice-versa, but the total energy remains the same so he would never stop.

In a real world, however, this is not true. In fact, in a real world some frictional force are present, in particular:

- friction: this force is due to the contact between the skateboard and the surface of the halfpipe, and its direction is always opposite to the motion of the skateboarder

- Air resistance: this force is due to the resistance opposed by the molecules of air that the skateboarder meets during his motion, and its direction is also opposite to the motion of the skateboarder

This two forces are said to be non-conservative forces, which means that they cause some of the mechanical energy of the skateboarder to be "lost", in the sense that it is dissipated as heat and it is no longer available for the skateboarder.

Therefore, the correct option is

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

7 0
3 years ago
In a closed system, a cart with a mass of 1.5 kg is rolling to the right at 1.4 m/s, while another cart of mass 1.0 kg is rollin
bija089 [108]

Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

Velocity of Cart B = 1.4 m/s towards left

Momentum (P)= Mass × Velocity

P_A=1.5 kg\times 1.4 m/s=2.1 kg m/s

P_B=1.0 kg\times (-1.4m/s)=-1.4 kg m/s

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)

Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

Hence, the correct option is (c.).

7 0
3 years ago
Read 2 more answers
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
4 years ago
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