All categories

3 years ago

how is acceleration related to force when mass is constant , according to Newton’s second law of motion

1 answer:

8 0

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

You might be interested in

a car is traveling 20 m/s. it takes 4 seconds with a 7,500 n force to come to a stop. what is the mass of the car

aev [14]

Answer:

mass equals 375 kg

Explanation:

f=ma

f=7500

a=20

m=f/a

m=7500/20

m=375

4 0

3 years ago

Students who embrace an innate mindset might view their poor performance in math as inevitable or beyond their control

PilotLPTM [1.2K]

Answer:

True

Explanation:

4 0

4 years ago

Read 2 more answers

Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ : Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

7 0

3 years ago

Part A: Determine the wavelength of photons that can be emitted

torisob [31]

Answer:

A λ = 97.23 nm

, B) λ = 486.2 nm

, C) λ = 53326 nm

Explanation:

With that problem let's use the Bohr model equation for the hydrogen atom

$E_{n}$ = -k e² /2a₀ 1/n²

For a transition between two states we have

$E_{nf}$ - $E_{no}$ = -k e² /2a₀ (1/ $n_{f}$ ² - 1 / n₀²)

Now this energy is given by the Planck equation

E = h f

And the speed of light is

c = λ f

Let's replace

h c / λ = - k e² /2a₀ (1 / $n_{f}$ ² - 1 / no₀²)

1 / λ = - k e² /2a₀ hc (1 / $n_{f}$ ² -1 / n₀²)

Where the constants are the Rydberg constant $R_{H}$ = 1.097 10⁷ m⁻¹

1 / λ = $R_{H}$ (1 / n₀² - 1 / nf²)

Now we can substitute the given values

Part A

Initial state n₀ = 1 to the final state $n_{f}$ = 4

1 / λ = 1.097 10⁷ (1/1 - 1/4²)

1 / λ = 1.0284 10⁷ m⁻¹

λ = 9.723 10⁻⁸ m

We reduce to nm

λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)

λ = 97.23 nm

Part B

Initial state n₀ = 2 final state $n_{f}$ = 4

1 / λ = 1.097 10⁷ (1/2² - 1/4²)

1 / λ = 0.2056 10⁻⁷ m

λ = 486.2 nm

Part C

Initial state n₀ = 3

1 / λ = 1,097 10⁷ (1/3² - 1/4²)

1 / λ = 5.3326 10⁵ m⁻¹

λ = 5.3326 10-5 m

λ = 53326 nm

5 0

3 years ago

In an experiment to estimate the acceleration due to gravity, a student drops a ball at a distance of 1 m above the floor. His l

anyanavicka [17]

Answer:

$9.45179\ m/s^2$

$s=4.725895t^2$

Explanation:

t = Time taken = 0.46

u = Initial velocity

v = Final velocity

s = Displacement = 1 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0\times 0.46+\frac{1}{2}\times a\times 0.46^2\\\Rightarrow a=\frac{1\times 2}{0.46^2}\\\Rightarrow a=9.45179\ m/s^2$

The acceleration due to gravity is 9.45179 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=\frac{1}{2}at^2\\\Rightarrow s=\dfrac{1}{2}9.45179t^2\\\Rightarrow s=4.725895t^2$

The function is $s=4.725895t^2$

3 0

3 years ago