A jogger runs 5 km in 0.6 h, then 9 km in 0.5 hr. What is the average speed of the jogger

beks73 [17]

New substances are formed by chemical reactions. When elements react together to form compounds their atoms join to other atoms using chemical bonds. For example, iron and sulfur react together to form a compound called iron sulfide. Hopefully this will help you decide...

5 0

2 years ago

A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the

tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0

2 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

Two identical 3.0-kg cubes are placed on a horizontal surface in

DedPeter [7]

The magnitude of the force exerted by the left cube on the right cube is 17.64N.

<h3>What is frictional force?</h3>

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.

Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.

From the equilibrium of forces in vertical direction

Normal force N= 2m x g

friction force f = μN =μ(2m)g

From the equilibrium of forces in horizontal direction

F₁ =ma =0

using Newton's third law of motion, we get

F₁ - f =0

F₁ =f = μ(2m)g

Put the values, we get

F₁ = 17.64N

Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.

Learn more about friction force.

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5 0

1 year ago

NVMMMMMMMMMMMMMMMMMMMM

sp2606 [1]

Huh??????????????????????????

8 0

3 years ago