Can u answer 4 and 5 for me

A skateboarder is standing at the top of a tall ramp waiting to begin a trip. The skateboarder has

Mnenie [13.5K]

Has a skateboard. your gonna have to give more details the. that just one .

7 0

3 years ago

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.

4 0

3 years ago

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

Dvinal [7]

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

3 0

3 years ago

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago

When a man walks across the carpet and reaches for the door knob, why does he get shocked? doorknob Question 3 options: There is

Mumz [18]

The imbalance of electric charges within or on the surface of a material or between materials is known as static electricity. When you grab a doorknob, electrons jump from your hand to the knob .which conducts electricity, giving you a shock.

<h3>What is static electricity?</h3>

The imbalance of electric charges within or on the surface of a material or between materials is known as static electricity.

The charge will persist until it can be moved away by an electric current or an electrical discharge.

when we walk on the carpet with socks our feet slide electrons off the carpet. it leads to leaving you with a slightly negative static charge.

When you grab a doorknob electrons jump from your hand to the knob which conducts electricity giving you a shock.

Due to the static electricity jump from your hand to the knob .which conducts electricity giving you a shock.

To know more about static electricity refer to the link ;

brainly.com/question/821085

3 0

2 years ago