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kolbaska11 [484]

3 years ago

6

Use the Pythagorean theorem to answer this question. A paper airplane is thrown westward at a rate of 6 m/s. The wind is blowing

at 8 m/s toward the north. What is the actual velocity of the airplane?
2 m/s, northwest
10 m/s, northwest
14 m/s, northwest
48 m/s, northwest

2 answers:

Goryan [66]3 years ago

4 0

6^2 + 8^2 = 36 + 64 = 100

sqrt(100) = 10 m/s northwest

Tema [17]3 years ago

4 0

The correct answer to the question is : 10 m/s , northwest.

EXPLANATION:

As per the question, the velocity of wind V = 8 m/s towards north.

The velocity of the paper airplane is V' = 6 m/s towards west.

We are asked to calculate the actual velocity of the airplane.

By putting Pythagorean theorem, the actual velocity of the airplane is calculated as -

$V_{net}^2 =\ V^2+V'^2$

⇒ $V_{net}^2=\ 6^2+8^2$

⇒ $V_{net}^2\ =\ 100$

⇒ $V_{net}=\ \sqrt{100}\ m/s$

⇒ $V_{net}=\ 10\ m/s$ towards northeast.

Hence, the correct answer is 10 m/s , northwest.

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Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0

3 years ago

Read 2 more answers

you are driving along a country road when you suddenly notice a log in the road ahead of you and immediately apply your brakes.

DiKsa [7]

a. The car will not hit the tractor.

b. The car would travel a distance of 52.07 m before stopping

c. At the moment when the car stopped, the tractor is 11.5 m in front of the car.

Linear motion

From the question, we are to determine of you will hit the tractor before you stop

First, we will determine the time it will take the car to stop

From the given information,

Initial velocity, u = 27.0 m/s

a = -7 m/s² (Negative sign indicates deceleration)

v = 0 m/s (Since the car will come to stop)

From one of the equations of linear motion,

v = u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration

and t is the time taken

Putting the parameters into the equation, we get

0 = 27 + (-7)t

7t = 27

t = 27/7

t = 3.857 secs

This is the time it will take the car to stop

Now, we will determine the distance the car would travel after applying the brakes

Using the formula

S = (u + v) / t

Where S is the distance traveled

S = [(27 + 0) / 2] * 3.857

S = 13.5 * 3.857

S = 52.0695 m

S ≅ 52.07 m

This means the car would travel 52.07 m after applying the brakes

Now, we will determine the distance the tractor would have traveled when the car came to a stop

Speed of the tractor = 10.0 m/s

Time taken for the car to stop = 3.857 secs

Using the formula,

Distance = Speed × Time

Distance = 10.0 × 3.857

Distance = 38.57 m

Now, we will determine the distance between the car and the tractor when the car finally stopped

Distance between the car and the tractor = Distance ahead + Distance traveled by the tractor after the car stopped - Distance traveled by car after applying the brakes

Distance between the car and the tractor = 25 m + 38.57 m - 52.07 m

Distance between the car and the tractor = 11.5 m

Therefore, the distance the tractor was ahead of the car + the distance the tractor traveled after the car stopped is more than the distance the car traveled after applying the brakes (25 m + 38.57 m > 52.07), the car will not hit the tractor.

To know more about distance, refer: brainly.com/question/10903482

#SPJ4

<u><em>[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tractor 25.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. you immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2 .</em></u>

<u><em>a.will you hit the tractor before you stop?</em></u>

<u><em>b.how far will you travel before you stop or collide with the tractor?</em></u>

<u><em>c.if you stop, how far is the tractor in front of you when you finally stop?]</em></u>

7 0

1 year ago

suppose we have two masses m1=2000 g and m2=4000g, where m1 is moving with initial velocity v1,i=24m/s and m2 is at rest at t=0s

Veseljchak [2.6K]

Answer:

The final velocity is 8 m/s and its direction is along the positive x-axis

Explanation:

Given :

Mass, m₁ = 2000 g = 2 kg

Mass, m₂ = 4000 g = 4 kg

Initial velocity of mass m₁, v₁ = 24i m/s

Initial velocity of mass m₂, v₂ = 0

According to the problem, after collision the two masses are stick together and moving with same velocity, that is, $v_{1f}$ .

Applying conservation of momentum,

Momentum before collision = Momentum after collision

$m_{1} v_{1} +m_{2} v_{2} =(m_{1}+m_{2}) v_{1f}$

Substitute the suitable values in the above equation.

$2\times24 +4\times0 =(2+4}) v_{1f}$

$v_{1f}=8i\ m/s$

8 0

3 years ago

Electric current in electric bulb is 20 mA, what is the charge flowing through the filament in 4 sec

anzhelika [568]

Answer:

The charge flowing through the filament is 0.08C

Explanation:

$I=20mA = 0.02A\\t= 4s\\$

Plug your given values into the charge and current formula:

$I=\frac{Q}{t} \\Q=I*t\\Q= (0.02A)(4s)\\Q= 0.08C$

6 0

3 years ago

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge

MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)

Keeping in mind the relationship between charge, voltage and capacitance:

$C=\frac{Q}{V}$

we can re-write (1) as

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

$C=4 \pi \epsilon_0 R$

where R is the radius of the sphere. Substituting this into (2), we find

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}$ (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

$R_2 = 2R_1$

So the eq.(3) becomes

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}$

And re-arranging it we find:

$Q_2 = 2Q_1$

And since we know that the total charge is

$Q_1 + Q_2 = 7.00 nC$

we find

$Q_1 = 2.33 nC\\Q_2 = 4.67 nC$

3 0

4 years ago