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exis [7]
3 years ago
6

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

You might be interested in
Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b
AleksandrR [38]

Answer:

  Ф = 2.179 eV

Explanation:

This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.

            E = K + Ф

the energy of the photons is given by the Planck relation

            E = h f

we substitute

           h f = K + Ф

           Ф= hf - K

the speed of light is related to wavelength and frequency

            c = λ f

            f = c /λ

            Φ = \frac{hc}{\lambda } - K

let's reduce the energy to the SI system

            K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J

calculate

           Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹  -1.424 10⁻¹⁹

           Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹

           Ф = 3.4571 10⁻¹⁹ J

         

we reduce to eV

           Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

           Ф = 2.179 eV

4 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
Why would an egg break immediately when it hits the ground?
Elza [17]

As an egg falls towards the floor, it begins to travel faster and faster. When it slams into the floor, the egg is stopped almost immediately. This force of the floor against the eggshell is too large, so it breaks.
6 0
3 years ago
Read 2 more answers
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
Highest density if electrostatic charge in a metal is found in​
arsen [322]
There’s no picture so how r we supposed to answer it
8 0
3 years ago
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