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storchak [24]
3 years ago
11

A block with mass m = 4.3 kg is attached to two springs with spring constants kleft = 35 N/m and kright = 48 N/m. The block is p

ulled a distance x = 0.23 m to the left of its equilibrium position and released from rest.What is the magnitude of the net force on the block (the moment it is released)?
Physics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

Explanation:

A restoring force is created when a spring is either stretched or compressed.

Restoring force created by first spring = spring constant x contraction

= 35 x .23

= 8.05 N .

It will act towards the right .

Restoring force created by second spring = spring constant x extension

= 48 x .23

= 11.04 N .

It will also act towards the right .

Total force created due to contraction in the first spring and extension in the right spring = 8.05  +11.04 N

= 19.09 N .

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Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
photoshop1234 [79]

Answer:

The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).

Explanation:

First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.

The equation for the position of the car is:

x = v · t

Where

x = position at time t

v = velocity

t = time

x = v · t

27 m = 17 m/s · t

27 m / 17 m/s = t

t = 1.6 s

Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:

The position of the bolt will be:

y = y0 + 1/2 · g · t²

Where

y = height of the bolt at time t

y0 = initial height of the bolt

g = acceleration due to gravity

t = time

Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:

y = y0 + 1/2 · g · t²

0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²

y0 = 13 m

8 0
3 years ago
What is induced current​
Brrunno [24]

Answer:

A current can be induced in a conducting loop if it is exposed to a changing magnetic field. ... In other words, if the applied magnetic field is increasing, the current in the wire will flow in such a way that the magnetic field that it generates around the wire will decrease the applied magnetic field.

Explanation:

8 0
2 years ago
Match these terms with the correct examples.
posledela
1. liquid solution to a. oceans
2. gaseous solution to b. clouds
Not sure about 3 and 4.
3 might be oxygen but I think that's 5. element.

Hope this helps, not sure about water and air though.
4 0
3 years ago
Read 2 more answers
At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is
n200080 [17]

a) 8.93\cdot 10^{-30} N, downward

b) 1.76\cdot 10^{-17} N, upward

c) 1.20\cdot 10^{-17} N, downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.8 m/s^2 is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

F=qE

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

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

E=110 N/C is the strength of the electric field

Substituting,

F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N

And since the electron has negative charge, the  direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

F=qvB

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

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

v=1.50\cdot 10^6 m/s is the velocity

B=50.0\mu T = 50.0 \cdot 10^{-6} T is the strength of the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0
2 years ago
A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12
Kryger [21]
Same formula as the last question. x = vt + (1/2)at^2. In this case, v = 0, t = 12, and a = 2.0. Plug in the values and solve for x (which is change in position) x = (0)(12) + (1/2)(2.0)(12^2) x = (1/2)(2.0)(144) x = (1)(144) x = 144 So the car will travel 144 meters in 12 seconds.
8 0
2 years ago
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