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Doss [256]
3 years ago
11

An organism that cannot make its own food called

Physics
2 answers:
Rzqust [24]3 years ago
8 0

Answer:

heterotroph

Explanation:

nevsk [136]3 years ago
5 0

Answer:

a predator cannot make its own food

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40 meters divided by 5
denis23 [38]

Answer:

8 meters

Explanation:

40/ 5 = 8

8 0
2 years ago
Please guys help me i will help u too ​
matrenka [14]

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calculate the cars acceleration usingv=u+at

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m/s. After 5 s the car reaches the bottome of the hill. Its speed at the bottom of the ... accelerating left a rownie. 10. A cart slows down while moving away from the ... does it need to accelerate to a velocity of 20 m/s

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3 years ago
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A restaurant records the number of tables served each night, and the results have the values: minimum = 3, lower quartile = 14,
dybincka [34]

To choose the correct box plot, verify each of the options and make sure all the values in the plot match the values provided.

<h3>How to identify the median?</h3>

In a box plot, this value is represented by a vertical line located in the middle of the graph.

<h3>How to identify the maximum and the minimum?</h3>

The maximum is the value located on the farthest right, while the minimum is located on the farthest left.

<h3>How to identify the quartiles?</h3>

Divide the graph into 4 and analyze how much each quartile represents.

Learn more about graphs in: brainly.com/question/16608196

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5 0
2 years ago
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v
zhenek [66]

Answer:

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4 0
3 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
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