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4 years ago

PEOPLE PLEASE HELP ASAP!!!-How does Force effect acceleration?

Physics

2 answers:

Lady_Fox [76]4 years ago

4 0

We know:

Force =Mass×Accerlation

$∴Accerlation = \dfrac{force}{mass}$

So by seeing this formula we came to know that accerlation is directly proportional to force and inversely proportional to mass.

So by increase of force acceleration also increases.

kaheart [24]4 years ago

3 0

Answer:

If the force is higher then the resitance o fht eobject then force will make object accelerate.

Explanation:

Does that make sense. If you need something else as an answer tell me in the comments and please mark with a heart and brainliest. If you don't mind

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How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th

Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

$R_t=\dfrac{r_1\times r_2}{r_1+r_2}$

Taking reciprocal of the above equation.

$\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}$

We can also write the above equation as :

$\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}$

On simplification of above equation,

$\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}$

$\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}$

Hence, proved.

4 0

3 years ago

Consider three capacitors C1, C2, and C3 and a battery. If

VLD [36.1K]

Answer:

Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC

Explanation:

Since the same voltage in the battery is used for the entire rundown,

From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",

Q = C₁V = 30 μC

V = (30/C₁)

the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC

The charge on both capacitors are the same and equal to 15 μC (because they are in series)

Q = (Ceq) V = 15 μC

(Ceq) = (15/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (15/V) μF

(Ceq) = 15 ÷ (30/C₁)

(Ceq) = 15 × (C₁/30) = 0.5 C₁

(1/Ceq) = (2/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₂)

(2/C₁) = (1/C₁) + (1/C₂)

(2/C₁) - (1/C₁) = (1/C₂)

(1/C₁) = (1/C₂)

C₁ = C₂

C₂ = C₁

C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.

The charge on both capacitors are the same and equal to 10 μC (because they are in series)

Q = (Ceq) V = 10 μC

(Ceq) = (10/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (10/V) μF

(Ceq) = 10 ÷ (30/C₁)

(Ceq) = 10 × (C₁/30) = 0.333 C₁

(1/Ceq) = (3/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₃)

(3/C₁) = (1/C₁) + (1/C₃)

(3/C₁) - (1/C₁) = (1/C₃)

(2/C₁) = (1/C₃)

C₁ = 2C₃

C₃ = (C₁/2)

C₁, C₂, and C₃ are connected in series with one another and

with the battery, what is the charge on C₁

The charge on C₁ is the same as the charge on all the capacitors and equal to Q,

Q = (Ceq) V

(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)

Substituting for C₂ and C₃

C₂ = C₁ and C₃ = (C₁/2)

(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)

(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)

(1/Ceq) = (4/C₁)

Ceq = (C₁/4)

Q = (Ceq) V = (C₁/4) V

But recall that V = (30/C₁) from the first connection

Q = (C₁/4) (30/C₁)

Q = (30/4) = 7.5 μC

Hope this helps!

6 0

3 years ago

Review

pickupchik [31]

Her average speed per hour, was 42.5 miles per hour

6 0

3 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or

Stolb23 [73]

Answer:

I = (1.80 × 10⁻¹⁰) A

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = (1.0 × 10⁻¹⁵) T

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 3.6 cm = 0.036 m

(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)

4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036

I = (1.80 × 10⁻¹⁰) A

Hope this Helps!!!

3 0

4 years ago