You pull your sled through the snow a distance of 500 meters with a horizontal force of 200 newtons. How much work did you do?

Answers

Answer 1
Answer:

The force applied to pull an object is 200 N and the distance it sleds through the snow is 500 meters, so the amount of work done over here is 1,00,000 Nm.

What is Work?

When an external force moves an object over a distance, at least a fraction of that force must be exerted in the displacement's direction. This is known as work in physics. By multiplying the length of the path by the component of the force acting along the path, work can be calculated if the force is constant. The work W is equal to the force f times the distance d, or W = f.d, to mathematically describe this idea.

According to the question, the given values are :

Distance, d = 500 meters

Force, f = 200 Newton

So the formula for work done= force× displacement

W = 200 × 500

W = 1,00,000 Nm

Hence, the work done over here is 1,00,000 Nm.

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Answer 2
Answer: the answer will be u need to divide 500 by 200 so a force of 150 ig

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Which coefficient is needed in front of NaNO3 to balancethe equation Na2S +Zn[NO3]2_ZnS+NaNO#

Answers

So Let's adjust the condition! So to ken how to compose what number of components are on each of the sides from this condition.  The coefficient has to require the apply that the entire aggravate that is denotes we have to decide what number of is on the reactants side so: Na=2 , O=6, N=2 So there is a subscript of 3, so we can now surmise that we can duplicate 2*3=6. So the final answer shall be 2.

An ionic bond form when atoms

Answers

between two ions of opposite charges. In ionic bonding, electrons transfer from one atom to another.

(x - 4)2 = 36 Which of the following expressions represents the solutions to the given equation?

Answers

X = 22 is what I think.

BY what factor does the kinetic energy of a moving body change when its speed is reduced to half? please help i have an exam and i need an answer

Answers

Answer:

Kinetic Energy = 1/2 mass*velocity^2

K=1/2mv^2

Therefore if you reduce the speed of an object by 1/2, K reduced to 1/4 its value.

A dolphin leaps out of the water at an angle of 36.6° above the horizontal. The horizontal component of the dolphin's velocity is 7.87 m/s. Calculate the magnitude of the vertical component of the velocity.

Answers

Answer:

Explanation:

Given

Dolphin leaps out an angle \theta =36.6^(\circ)

Horizontal component of dolphin velocity u_x=7.87\ m/s

Suppose u is the launch velocity of dolphin

therefore u\cos \theta =u_x---1

and vertical velocity u_y=u\sin \theta ----2

divide 1 and 2 we get

\tan \theta =(u_y)/(u_x)

u_y=u_x\tan \theta

u_y=7.87\cdot \tan (36.6)

u_y=5.84\ m/s

How does the density of the metal in the crushed can compare with the metals density before the can was Crushed

Answers

dens = mass/volume. after crushing, vol less so density more. the actual density of the metal is the same, density of can changes

Final answer:

The density of a metal can, whether it is crushed or not, remains the same because density is an intrinsic property of matter, dependent only on the mass and volume of the substance. Changing the can's shape alters its volume but does not affect its mass or its density.

Explanation:

The density of a substance is the ratio of its mass to its volume (Density = Mass / Volume). This physical property is intrinsic to the substance, meaning it doesn't change regardless of the quantity or form the material is in.

When you crush a metal can, you're changing its shape and thereby altering its volume. However, the mass remains the same because you haven't removed or added any material. As a result, the density of the metal in the crushed can stays the same as the metal's density before the can was crushed.

Let's consider a practical example: a block of brass and a piece of rebar. Both have the same mass and are made of the same metal, but the block of brass has a larger volume. Nevertheless, the densities of these substances remain constant. That's because density is an intrinsic property, reflecting the compactness of material in a substance. Hence, even though you change the can's shape (thus, its volume), you have no impact on the metal's density.

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