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2 years ago

If an object accelerating at −1.5m/s^2 takes 1.2s to reach 5.0m/s, what was its initial speed?

Physics

1 answer:

USPshnik [31]2 years ago

3 0

-1.5 m/s^2 x 1.2 seconds = -1.8 m/s

It is a negative value which means the object slowed down. The object would have originally been going that amount more.

5.0 + 1.8 = 6.8 m/s

answer: 6.8 m/s

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puteri [66]

Answer:

The answer is proteins

4 0

3 years ago

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in

Ilya [14]

Answer:

F = 2π I R B

Explanation:

The magnetic force is described by the equation.

F = q v x B = i L x B

Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

F = i L B sin θ

In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

F = i L B

The loop length is

L = 2π R

F = i 2π R B

F = 2π I R B

The force is in the loop

8 0

3 years ago

Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a

babymother [125]

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

8 0

3 years ago

Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two road

Gwar [14]

Explanation:

It is given that,

$\frac{dx}{dt}$ = -40 mi/h, $\frac{dx}{dt}$ = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find $\frac{dz}{dt}$ . Equation for the given variables according to the Pythagoras theorem is as follows.

$z^{2} = x^{2} + y^{2}$

Now, we will differentiate each side w.r.t 't' as follows.

$2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

or, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

= $\frac{1}{5}(4 \times (-40) + 3 \times (-40))$

= $\frac{-140 - 120}{5}$

= 52

Thus, we can conclude that the cars are approaching at a rate of 52 mi/h.

7 0

3 years ago

A 400-kg object has a 100-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

aliya0001 [1]

Answer:

The answer to your question is a = 0.25 m/s²

Explanation:

Data

mass = m = 400 kg

Force = F = 100 N

acceleration = a = ? m/s²

Process

To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.

Formula

F = ma

solve for a

a = $\frac{F}{m}$

Substitution

$a = \frac{100}{400}$

Simplification and result

a = 0.25 m/s²

5 0

3 years ago