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kompoz [17]
3 years ago
13

Austin buys a new moped he travel 3 km south and then 4 km east and then 3km north how far does he need to go to get back to whe

re he started
Physics
1 answer:
Ira Lisetskai [31]3 years ago
4 0
The first thing to do is to define the origin of the coordinate system as the point at which the moped journey begins.
 Then, you must write the position vector:
 r = -3j + 4i + 3j
 Rewriting
 r = 4i
 To go back to where you started, you must go
 d = -4i
 That is to say, must travel a distance of 4Km to the west.
 Answer
 West, 4km.
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A parked car begins to roll down a hill, what can you conclude from that observation?
Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
The weight of an object is measured in air to be 7.0 N. The
Vinil7 [7]

Answer:

Buoyant force = 3.0 N

The object will not float.

Explanation:

Apparent weight of a body immersed in water is the actual weight of object minus buoyant force

Given in the question that;

Weight of object in air = 7.0 N

Apparent weight of object = 4.0 N

4.0 N = 7.0 N - Buoyant force

Buoyant force = 7.0 - 4.0 = 3.0 N

In this case, the buoyant force is less than weight of the object thus the object will sink.

5 0
3 years ago
When energy is added to a wave, how can the wave change?.
nadya68 [22]

Answer:

When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

6 0
2 years ago
What does buoyancy depend on 
Marta_Voda [28]
... the density of the liquid
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8 0
4 years ago
Read 2 more answers
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume
Svetach [21]

Answer:

9.56\cdot 10^{-7} C

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

A=L^2=(0.19)^2=0.036m^2

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

E=3\cdot 10^6 N/C

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C

7 0
4 years ago
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