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grin007 [14]
3 years ago
7

Convert 1,265,341 mm to m.

Physics
2 answers:
kvasek [131]3 years ago
8 0
It’s easy the answer is 1265.341 I think so because It was online :)
Thepotemich [5.8K]3 years ago
7 0
The answer in Meters is going to to 1265.341
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Match the organs to the functions they perform during digestion.<br> passes food to esophagus
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Answer:

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Soloha48 [4]
Answer: A
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Express 50,000 m/s2 in scientific notation. 5.0 x 103m/s2 0.5 x 105m/s2 5.0 x 104m/s2
Tresset [83]
Answer would be 5.0x104m/s2
8 0
3 years ago
Read 2 more answers
A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.
NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:  

- A uniform motion with constant velocity along the horizontal direction  

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction  

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

s=u_y t+\frac{1}{2}at^2

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

u_y=0 is the initial vertical velocity  (the missile is thrown horizontally)

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

v_x = 45 m/s

Therefore, the distance covered in a time t is

d=v_x t

and by substituting t = 3.91 s, we find the horizontal range of the missile:

d=(45)(3.91)=176 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0
3 years ago
How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr
Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

Q =mc\Delta T

where,

m = mass

c = specific heat

\Delta T = Change in temperature

Therefore the total heat exchange is given as

\Delta Q = Q_w+Q_v

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

Our values are given as,

Total mass is M_T = 200lb ,however the mass of solid vegetable and water is given as,

m_v= 0.4*200lb = 80lb

m_w=0.6*200lb=120lb

T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF

Replacing at our equation we have,

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)

\Delta Q = 22411.2Btu

Therefore the heat removed is 22411.2 Btu

6 0
3 years ago
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