Which of the following is an example of a luminous object? <br>Dog<br> Tree <br>Candle <br>Moon

HELP ;

mestny [16]

Answer:

The frequency would double.

Explanation:

Given:

Speed of wave (v) = constant.

Frequency of wave initially (f₁) = 2 Hz

Initial wavelength of the wave (λ₁) = 1 m

Final wavelength of the wave (λ₂) = 0.5 m

Final frequency of the wave (f₂) = ?

We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.

Therefore, framing in equation form, we have:

Wavelength × Frequency = Speed

$\lambda\times f=v$

It is given that speed of the wave remains the same. So, the product must always be a constant.

Therefore,

$\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2$

Now, plug in the given values and solve for 'f₂'. This gives,

$1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz$

Therefore, the final frequency is 4 Hz which is double of the initial frequency.

f₂ = 2f₁ = 2 × 2 = 4 Hz

So, the second option is correct.

7 0

3 years ago

Read 2 more answers

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just

bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

2 years ago

Read 2 more answers

When you need to pass a motorcycle, _____________.A. move into the left lane to pass the motorcycle B. pass to the left using th

erma4kov [3.2K]

Answer:

A) move into the left lane to pass the motorcycle

Explanation:

According to law, when it is needed to pass other vehicles, it requires you to only pass other vehicles on the left (using the left lane).

When passing a motorcyclist, remember to give him/her the same full lane width as other vehicles. Never drive in the same lane with a motorcyclist, even if the lane is wide enough to fit your vehicle and the motorcyclist.

5 0

3 years ago

Which of the following is an example of a luminous object? Dog Tree Candle Moon​

Which of the following is an example of a luminous object? Dog Tree Candle Moon