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3 years ago

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A student squeezes a clothespin 82 times in a minute. Then, using the same hand and the same clothespin, he squeezes the clothes

pin 68 times in a minute. State one biological reason for the decrease in the number of squeezes during the second trial.

1 answer:

tia_tia [17]3 years ago

3 0

Could have gotten tired after the first trial. ? i’m not sure if that’s it

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A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

3 0

3 years ago

What is the force required to move a block of mass 150 pound by a distance of 5ft in 8 seconds?

Nadusha1986 [10]

Force required to move a block is 1.615 N

Given:

mass of block = m = 150 pounds = 68 kg

distance = d = 5 ft = 1.52 metres

time = t = 8 sec

To Find:

force required to move the block

Solution: Force is defined as product of mass and acceleration and it's unit is N or Newton.

Velocity = displacement/ time = 1.52 / 8 = 0.19 m/s

Acceleration = velocity/time = 0.19/8 =

0.023 m/s^2

Force = mass x acceleration = 68x0.023 = 1.615 N

Hence, force required to move the block is 1.615 N

Learn more about Force here:

brainly.com/question/12970081

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8 0

1 year ago

Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force F, where r is the dista

Marrrta [24]

Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

W = ∫ F.dr

Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product

W = ∫ F dr

W = G m₁m₂ ∫ 1 /r² dr

W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

W = G m₁m₂ (-1 / Re + 1 / ∞)

W = - G m₁ m₂ / Re

Let's calculate

W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶

W = - 5.01 10¹⁰ J

4 0

3 years ago

power of a crane is 25000 watt calculate the time required by it to lift a load of 6000 kg up tp the height of 20 m

saul85 [17]

Solution:

We have,

Power [P] = 25000 Watt

Mass [m] = 6000 kg

Height [h] = 20 metres

Time [t] = ?

Now,

P = W/t = F x d/t = mxgx h/t

Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10

m/s^2]

Or, t = 6000 x 10 x 20/25000

Or, t = 1200/25

Therefore, t = 48 second

Hence, the required time for the crane to lift the load is 48 seconds.

8 0

3 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago