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3 years ago

You are riding in an elevator on the way to the

Physics

1 answer:

timofeeve [1]3 years ago

4 0

Answer:

F = 104.832 N

Explanation:

given,

upward acceleration of the lift = 1.90 m/s²

mass of box containing new computer = 28 kg.

coefficient of friction = 0.32

magnitude of force = ?

box is moving at constant speed hence acceleration will be zero.

Now force acting due to lift moving upward =

F = μ m ( g + a )

F = 0.32 × 28 × ( 9.8 + 1.9 )

F = 104.832 N

hence, the force applied should be equal to 104.832 N

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English units of distance include the

Masja [62]

Answer:mile

Explanation: heres a hint think aboyt the distance between your house to school

8 0

3 years ago

A ball of mass 6 kg is moving to the right with a velocity of 4 m/s when it strikes a 12 kg green block moving to the left at 5

Marat540 [252]

Answer:

Velocity of both masses after the collisio

Explanation:

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3 years ago

La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

$\lambda=3.33 cm = 0.033 m$ is the wavelength

So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

8 0

3 years ago

The three factors that determine the amount of potential energy in an object are?

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Answer:

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4 0

3 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

3 years ago