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3 years ago

Define what is false

Physics

1 answer:

GalinKa [24]3 years ago

3 0

Answer: not according with truth or fact/ incorrect.

Explanation:

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45. When you park on a level road next to a curb: A. Your wheels must be within 18 inches of the curb B. Your front wheels must

tekilochka [14]

Answer:

Explanation:

You must be within 18 inches.

4 0

2 years ago

After a hold up, Robin Banks flees in his 1575 kg getaway car at 20 m/s. He crashes into a 45 kg highway barrel which is at rest

fiasKO [112]

Answer:

v = 38.9 m/s

Explanation:

7 0

3 years ago

Read 2 more answers

You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee

Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour

New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.

Hope this Helps!!!

7 0

3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago

Thallium consists of 29.5%TI-203 and 70.5%TI-205. What is the relative atomic mass of thallium?

Kazeer [188]

Since you are referring to the TI-203 and TI-205, you need to know the actual masses of these two isotopes. TI-203 has 202.9723 amu and TI-205 has 204.9744 amu. Since you are concluding that this Thallium have 29.5% (Ti-203) and 70.5% (Ti-205), you need to multiply the percentage to the actual masses of the isotopes. With that, you should be able to get 204.3833 amu

7 0

2 years ago

Define what is false​

Define what is false