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sergeinik [125]
3 years ago
8

A race car circles 10 times around a circle with a diameter of 7 km track in 50 min. Using SI

Physics
1 answer:
vodka [1.7K]3 years ago
4 0

Average Speed = 23.33 m/s

<u>Explanation:</u>

Given:

Diameter = 7 km

So, to find the distance, we need to multiply it by 10 as the race car circles 10 times round this 7 km track.

Distance = 7 × 10 = 70 Km or 70,000 m

Time        = 50 mins

                = 50 × 60 = 3000 s

Now, in order to find the average speed for 10 cycles, we use the formula:

Speed = \frac {Distance} {Time}

So by substituting the above values in the given formula, we get:

Speed = \frac {70,000} {3000}

Speed = 23.33 m/s

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The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
35. (II) A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging on a 2.8-m long string, which makes th
pentagon [3]

Answer:

Explanation:

Let mass of bullet = m1 = 28g= 0.028 kg

mass of pendulum = m2 = 3.1 kg

8 0
3 years ago
A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and
Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ <em>F</em> = <em>N</em> - <em>W</em> = 0

• net horizontal force:

∑ <em>F</em> = <em>Fs</em> = <em>m a</em>

where

<em>N</em> = magnitude of normal force

<em>W</em> = car's weight

<em>Fs</em> = mag. of static friction

<em>m</em> = car's mass

<em>a</em> = <em>v</em> ²/<em>R</em> = mag. of the centripetal acceleration

<em>v</em> = car's speed

<em>R</em> = radius of curve

Now,

• compute the car's weight:

<em>W</em> = <em>m g</em> = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

<em>N</em> = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

<em>Fs</em> = 0.5 <em>N</em> = 7350 N

• solve for the (maximum) acceleration:

7350 <em>N</em> = (1500 kg) <em>a</em>   →   <em>a</em> = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = <em>v</em> ²/ (35 m)   →   <em>v</em> ≈ 13 m/s

4 0
3 years ago
Air is compressed in a cylinder such that the volume changes from 100.0 to 10.0 in^3. The initial pressure is 50.0 psia and the
EleoNora [17]

Answer:

5953.42 J

Explanation:

Given:

Initial volume, V_i= 100 in³

Final Volume, V_f = 10 in³

Initial pressure = 50 psia

Temperature = 100° F = 310.93 K

For isothermal reversible process, work done is given as:

Work done = -2.303RTlog_{10}\frac{V_f}{V_i}

Where,

R is the ideal gas constant = 8.314 J/mol.K

or

Work done = -2.303\times8.314\times310.93log_{10}\frac{10}{100}

or

Work done = 5953.42 J

7 0
3 years ago
About how much of Earth's land surface is used for agriculture?<br><br> 8%<br> 18%<br> 28%<br> 38%
Alexxandr [17]
We use about 38 percent of Earth's land surface for agriculture
6 0
3 years ago
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