The velocity of a motor car moving along a road increases from 10m/s to 50{1} ms in 8s. Find its avarege acceleration. RESULT

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

a

$\theta = 23.32^o$

b

$\mu_s = 0.27$

c

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

So

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

so

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

4 0

3 years ago

Two 2.0 cm by 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires of the terminals of a 9.0 V battery.

ale4655 [162]

Answer:

Explanation:

Area of electrodes, A = 2 cm x 2 cm = 4 cm²

Separation between electrodes, d = 1 mm

Voltage, V = 9 V

(a)

Let C is the capacitance between the electrodes

$C = \frac{\epsilon _{0}A}{d}$

$C = \frac{8.854\times 10^{-12}\times 4\times 10^{-4}}{1\times 10^{-3}}$

C = 3.54 x 10^-12 F

Let q be the charge on each of the electrode

q = C x V

q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C

(b)

As, the battery is disconnected the charge on the electrodes remains same.

(c)

As the battery is connected the voltage is same.

capacitance is change.

As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C

6 0

3 years ago

An electric current flows through a parallel circuit. Which units tell you about the amount of electrons traveling in the circui

kompoz [17]

'Ampere' is the unit of current. That's the rate at which
electrons travel in the circuit ... the number of electrons
every second. If you wanted the actual amount or number
of electrons, you'd need to know the length of time too.

It doesn't matter whether we're talking about a parallel or
series circuit.

8 0

3 years ago

11. A fundamental property of light is that it: 15

myrzilka [38]

Answer:

Curves around objects

Explanation:

Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.

5 0

2 years ago

Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.

iren [92.7K]

Answer:

F_n = 5.65E-11 N

d = 1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

$F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}$

For $F_n$

$F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}$

Dividing the above equations we get

$\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N$

F_n = 5.65E-11 N

$F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m$

d = 1.20682E-31 m

8 0

3 years ago