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shtirl [24]
3 years ago
8

4. If you swing wide to the left before turning right, another driver may try to pass you on the right. True or False

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

True.

Explanation:

Don't turn wide to the left as you start the turn. A driver behind may think you are turning left and try to pass you on the right. You may crash into the other vehicle as you complete your turn.

Instead, slowly give yourself and others more time to avoid problems, keep the rear of the vehicle close to the curb. This will stop other drivers from passing you on the right. This is called (button Hook)

If you are driving a truck or bus that cannot make the right turn without swinging into the other lane, turn wide as you complete the turn.

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A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0
Kazeer [188]

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

3 0
3 years ago
A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc
Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

7 0
3 years ago
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A person takes a trip, driving with a constant speed of 89.5km/h, except for a 22.0−min rest stop. If the person's average speed
Elanso [62]

Answer:

stop using brainly and learn this bs in class

Explanation:

8 0
4 years ago
Which of the following is a correct comparison of a compact fluorescent lamp and incandescent light bulbs?
Alekssandra [29.7K]
The correct answer is C.  

Hope I helped.  
8 0
3 years ago
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One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What
balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, f_1 (first harmonic):

f_{n+1}-f_n = f_1 (1)

In this problem, we are told the frequencies of two successive harmonics:

f_n = 576 Hz\\f_{n+1}=648 Hz

So the fundamental frequency is:

f_1 = 648 Hz-576 Hz=72 Hz

Now we know that one of the the harmonics is f_n=216 Hz, so its next highest harmonic will have a frequency of

f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

f_n=n f_1 (2)

Since we know f_n = 288 Hz, we can solve (2) to find the number n of this harmonic:

n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

f_n=(2n+1) f_1 (2)

so, the difference between any two harmonics tube is equal to:

f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1 (1)

In this problem, we are told the frequencies of two successive harmonics:

f_n = 4699 Hz\\f_{n+1}=4953 Hz

So, according to (1), the fundamental frequency is equal to half of this difference:

f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz

Now we know that one of the harmonics is f_n=4191 Hz, so its next highest harmonic will have a frequency of

f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

f_n=(2n+1) f_1 (2)

Since we know f_n = 4445 Hz, we can solve (2) to find the number n of this harmonic:

n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17

7 0
3 years ago
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