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2 years ago

And I need help with seven and eight only I will appreciate it

Physics

1 answer:

Mumz [18]2 years ago

8 0

Answer:

7] Force = mass × acceleration

Force = 2 × 5

Force = 10 N

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

Velocity = 117.6 m/s

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A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera

Mama L [17]

Angular acceleration = (change in angular speed) / (time for the change)

change in angular speed = (zero - 2,600 RPM) = -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = -(4 1/3) rev / sec²

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = (216 2/3) revs

6 0

4 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

What part of the atmosphere contains dust, water vapor, and 75% of all gases

Alexeev081 [22]

Answer:

Troposphere

Explanation:

Our atmosphere yay

7 0

3 years ago

Read 2 more answers

The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A

vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

$C = \frac{Q}{V} (1)$

For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

$C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)$

So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

$Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)$

6 0

3 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Salsk061 [2.6K]

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

3 0

3 years ago