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3 years ago

14

A bicycle with an initial velocity of +6 m/s accelerates at a rate of +2 m/s2 for 3 seconds. What distance does the bicycle trav

el during this time?

1 answer:

irga5000 [103]3 years ago

3 0

Answer:

27 m

Explanation:

Given:

v₀ = 6 m/s

a = 2 m/s²

t = 3 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (6 m/s) (3 s) + ½ (2 m/s²) (3 s)²

Δx = 27 m

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Why are marble and granite used so often as a building materials?

Katyanochek1 [597]

Answer:

Granite is an excellent building material that provides some outstanding interior design opportunities whilst being incredibly sturdy and reliable. It has become one of the most popular building materials in modern construction, from kitchen surfaces through to paving slabs.You Can Expect Durability and Longevity. Firstly, marble is so popular around the world because of its durability in a wide variety of weather conditions. Structures that are over hundreds of years old made from marble are still standing to this day, and look as pristine as they day they were crafted.

Explanation:

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3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

3 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

3 years ago

This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ

soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

We notice that:

The kinetic energy is directly proportional to the mass
The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

B) The object's velocity doubled.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0

3 years ago