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2 years ago

300,000,000m/s × 2 = —— m

Physics

1 answer:

nekit [7.7K]2 years ago

7 0

Answer:

6 x 10⁸ m

Explanation:

300,000,000 m/s x 2 s

= 3 x 10⁸ x 2

= 6 x 10⁸ m

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a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0

2 years ago

How do you know if you have all the forces needed for a FBD?

Korolek [52]

It’s so many FBD meaning be more specific so we can help you

5 0

2 years ago

A train accelerates to a velocity of 500 m/s over time of 2s. The acceleration it experienced was 50m/s2. What was its initial v

ValentinkaMS [17]

Explanation:

We know ,

v = u + at
u = v - at
u = 500 - 2*50
u = 400 m/s

7 0

2 years ago

Determine the angular speed, in rad/s of:

ryzh [129]

Answer:

Explanation:

A. The earth about its axis:

The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

$\frac{1 rev}{24hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s$

B. The minute hand of the clock makes one revolution in 60 minutes

To convert this to rad per second, we have

$\frac{1 rev}{60mins}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 1.745 \times10^-3rad/s$

C. The hour hand of a clock completes one revolution in 12 hours

$\frac{1 rev}{12hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 1.454 \times10^-4 rad/s$

D. an egg beater turning at 300 rpm.

$\frac{300 rev}{1minute}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s=0.0218rad/s$

6 0

2 years ago

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

F = Mg

⇒F = 84.8×9.81

⇒F = 831.888 N

Area of the contact point

A = πR²

⇒A = π0.006²

⇒A = π0.000036 m²

Area of the four points is

4A = 0.000144π m²

Pressure

$p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa$

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0

3 years ago