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3 years ago

11

What is the force required to produce an acceleration of 46.4

1 answer:

Brilliant_brown [7]3 years ago

7 0

That depends on the mass of the object, and the unit of the '46.4' .

If the '46.4' is ' meters per second² ' , then the force required is

(mass of the object in kilograms) x (46.4) newtons .

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A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet

kogti [31]

The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.

4 0

3 years ago

Read 2 more answers

2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g

Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

Pressure = 235,200 Pa

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3 years ago

the inertial mass of an object is measured by exerting a force on the object and measuring the object's BLANK using an inertial

JulijaS [17]

Answer:

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3 years ago

You accidentally drop an eraser out of the window of an apartment 15 m above the ground

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Answer:

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3 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

3 years ago