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2 years ago

If a car starts at rest and accelerates at 3.4 m/s^2 over a distance of 200 meters, how fast will it be travelling at the end?

Physics

1 answer:

CaHeK987 [17]2 years ago

8 0

Answer:

36.9 m/s

Explanation:

From;

v^2 = u^2 + 2as

Where;

v = final velocity =?

a = acceleration = 3.4 m/s^2

u = initial velocity = 0 m/s

s = distance covered = 200 meters

v^2 = 0^2 + 2 * 3.4 * 200

v^2 = 1360

v = √1360

v = 36.9 m/s

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Interstellar matter is composed of ____________ found between stars in a galaxy.

Illusion [34]

This should be dust and gas

4 0

3 years ago

A projectile is fired horizontally off the top of a cliff with an initial velocity of 30 m/s. It hits the ground 2.0 seconds lat

Mandarinka [93]

Answer:

The horizontal distance traveled by the projectile is 60 m

Explanation:

Given;

initial horizontal velocity of the projectile, Vₓ = 30 m/s

time of the motion of the projectile, t = 2 s

The horizontal distance traveled by the projectile is given by the range of the projection;

X = Vₓt

X = 30 x 2

X = 60 m

Therefore, the horizontal distance traveled by the projectile is 60 m

Therefoe

6 0

2 years ago

A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. Part A) At the instant

Gre4nikov [31]

A) $2.2\cdot 10^{-19} N$

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

In this situation,

I = 5.20 A

r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

$B=\frac{(4\pi \cdot 10^7)(5.20)}{2\pi (0.046}=2.26\cdot 10^{-5} T$

The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

$F=qvB$

where

q is the magnitude of the charge of the particle

v is its velocity

B is the magnetic flux density

For the electron of this problem,

$q=1.6\cdot 10^{-19} C$ is the charge

$v=6.10\cdot 10^4 m/s$ is the speed

$B=2.26\cdot 10^{-5} T$ is the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(6.10\cdot 10^4)(2.26\cdot 10^{-5})=2.2\cdot 10^{-19} N$

B) Same direction as the current in the wire

First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:

- The thumb gives the direction of the current -> upward

- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)

Now the direction of the force can be found by using the right-hand rule. We have:

- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)

- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

8 0

3 years ago

An office window has dimensions 3.5 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.96 at

Nina [5.8K]

Answer:

Force=29789.55 N

Explanation:

Given data

dimension=3.5m by 2.1m

P(outside pressure)=0.96 atm

P(inside pressure)=1.00 atm

Force=?

Solution

Area=L×W

A=(3.5)×(2.1)

$A=7.35m^{2}$

As we know

F=PA

For P(pressure)

P=P(inside)-P(outside)

P=1.00-0.96

P=0.04 atm

convert atm to pascal (pa) we get

P=4053 pa

F=PA

F=(4053)×(7.35)

F=29789.55 N

8 0

3 years ago

A spring scale reads 20 N as it pulls a 4 kilogram box across a table. What is the magnitude of the force exerted by the mass on

Ivan

The answer is 40!!!!!!!!!!!!!

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3 years ago