How much work is done when a 5 N force moves a block 4 m?

What is the force due to gravity of a 38 kg student?

alukav5142 [94]

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

6 0

3 years ago

My mom and I walked 3,600 m in 90 minutes. What was our speed in m/min?

PtichkaEL [24]

Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.

40 mi/min

3 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

2 years ago

How do altitude temperature and air pressure relate

kiruha [24]

They relate because the further up you go, the colder it gets, and the air pressure decreases the further up you go. The altitude temperature and the air pressure both decrease, and that is their relationship. Altitude temperature decreases, the higher you go, and air pressure also decreases, the higher up you go. Therefore, the lower down you go, the higher the air pressure, and the higher the altitude temperature.

Hope this helps and have a nice day:)

7 0

2 years ago

Read 2 more answers

A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz

uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3) T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

λ = 2L / n n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

λ = 2 0.62 / 3

λ = 0.413 m

2) the velocity related to wavelength and frequency

v = λ f

v = 0.413 61

v = 25,213 m / s

3) let's use the equation

v = √T /μ

T = v² μ

T = 25,213² 3.4 10⁻⁴

T = 0.216 N

4) the rope tension is proportional to the hanging weight

T-W = 0

T = W

W = m g

m = W / g

m = 0.216 / 9.8

m = 22.04 10-3 kg

5) n = 2

λ = 2 0.62 / 2

λ = 0.62 m

6) v = λ f

v = 0.62 61

v = 37.82 m / s

7) T = v² μ

T = 37.82² 3.4 10⁻⁴

T = 0.486 N

8) m = W / g

m = 0.486 / 9.8

m = 49.62 10⁻³ kg

9) n = 1

λ = 2 0.62

λ = 1.24 m

v = 1.24 61

v = 75.64 m / s

T = v² miu

T = 75.64² 3.4 10⁻⁴

T = 2.572 10⁻² N

m = 2.572 10⁻² / 9.8

m = 262.4 10⁻³ kg

5 0

3 years ago