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1 month ago

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

1 answer:

6 0

The force exerted on the car during the stop measures 6975 N. Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = $\sqrt{2I/ce_{0} }$

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

1 month ago

Which of the following equations correctly expresses the relation between vectors A⃗ , B⃗ , C⃗ , and D⃗ shown in the figure?

Sav [3153]

C) B ⃗ = A ⃗ + C ⃗ Pay attention to the direction of the arrows (vectors).

7 0

1 month ago

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a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

Sav [3153]

To find the mass using a weight of 1.4 N:
1.4/9.8 = 0.1428 kg
The momentum is calculated as 0.1428 multiplied by 44.7, which is 6.38 kgm/s.

3 0

3 months ago

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A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

Softa [3030]

Explanation:

Here is the provided information:

Mass of oxygen contained = 100 mg = $100 \times 10^{-3}$ g

Thus, the moles of oxygen can be calculated using the following formula:

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of the cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Next, we can determine the cross-sectional area (A) as follows:

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of the tube = 11 cm = 0.11 m

Thus, the volume (V) is given by $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

For our calculations, we'll assume the internal pressure is P.

Moreover, $P_{atm}$ = 100 kPa = 100000 Pa,

The pressure difference amounts to 100000 - P

Thus, the force required to open is calculated as follows:

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are provided with the information that the force equals 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

By solving, we find,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

Using the ideal gas equation, PV = nRT

Hence, we will input the values into the equation as follows:

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Consequently, we can deduce that the gas temperature is 462.66 K.

6 0

1 month ago

A particle has a charge of -4.25 nC.

kicyunya [3294]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

The electric field can be calculated using

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The calculated magnitude is 611.32 N/C

The electric field direction is vertically downward due to the negative charge.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field source is 1.71436 m

4 0

3 months ago

Read 2 more answers