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MatroZZZ [7]
2 years ago
10

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni

tude and direction of a) A + B and b) A - B. Specify the directions relative due west.​
Physics
1 answer:
ozzi2 years ago
4 0

Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , A=-63i and B=63j

Now , A + B is given by :

A+B=-63i+63j

| A+B | = 63\sqrt{2}

Direction of A+B is 45° north of west .

Also , for A-B :

A-B=-63i-63j

|A-B|=63\sqrt{2}

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

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An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm
Rus_ich [418]

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

6 0
3 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
3 years ago
The object represented by this graph is moving
Vinil7 [7]

The object represented by this graph is moving toward the origin at constant velocity.

Option 3.

<u>Explanation:</u>

In the figure, x-axis is representing increase in the time and y-axis is presenting increase in the distance from bottom to up. But the line in the graph which is plotted is decreasing from high distance to small distance with increase in time. So this indicates that as the time is increasing, the distance is decreasing.

And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.

4 0
3 years ago
Read 2 more answers
3. When encountering low visibility from rain or fog, you
zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

When encountering low visibility from rain or fog, use your low beams and  fog lights. High beams will only increase the glare. If you can't see at least five seconds in front of you, don't drive. Pull over and put your hazards on until it clears up.

i just got the answer wrong and the drivers ed gave me this explanation !!

8 0
3 years ago
Read 2 more answers
If the spring constant is doubled , what value does the period have for a mass on a spring?
zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}

So the correct answer is

D. The period would decrease by sqrt (2)

8 0
3 years ago
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