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4 years ago

13

Can someone please help me. idk what to do

1 answer:

Alex_Xolod [135]4 years ago

3 0

Answer:

-4.5 m/s

Explanation:

Even without doing calculations, we know that a projectile has the same speed coming down as it did going up at the same height. Since it went up at 4.5 m/s, it returns down at -4.5 m/s.

To prove it with math, use the given information:

v₀ = 4.5 m/s

a = -9.8 m/s²

Δy = 0 m

Find: v

v² = v₀² + 2aΔy

v² = (4.5 m/s)² + 2 (-9.8 m/s²) (0 m)

v = -4.5 m/s

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What is the prefix notation of 0.0000738?

MrMuchimi

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

4 0

3 years ago

Read 2 more answers

Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physicallyrealistic. T

nata0808 [166]

Answer:

a. No, typical small pickup truck springs are not large enough to compress 0.61{\rm m}.

Explanation:

8 0

3 years ago

A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate

Sladkaya [172]

Answer:

$P_2=4091\ KPa$

Explanation:

Given that

T₁ = 290 K

P₁ = 100 KPa

Power P =5.5 KW

mass flow rate

$\dot{m}= 0.01\ kg/s$

Lets take the exit temperature = T₂

We know that

$P=\dot{m}\ C_p (T_2-T_1)$

$5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K$

If we assume that process inside the compressor is adiabatic then we can say that

$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}$

$\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\$

$2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}$

$P_2=40.91\times 100 \ KPa$

$P_2=4091\ KPa$

That is why the exit pressure will be 4091 KPa.

4 0

3 years ago

What happens at<br> the Bernoulli effect

Paul [167]

Answer:

Within a horizontal flow fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.

Explanation:

5 0

3 years ago

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h

Amiraneli [1.4K]

Answer:

Explanation:

a) Total distance traveled by the whale = 6.9 km + 1.8 km + 3.7 km = 12.4 km

b) the magnitude and the direction of the displacement of the whale = 6.9 km due east - 1.8 km due west = 5.1 due east

it final displacement = 5.1 km due east + 3.7 k due east = 8.8 km due east

c) average speed in km / hr = total distance traveled / total time = 12.4 km / 0.5 hr ( 30 min/ 60 min × 1 hr) = 24.8 km / hr

d) average velocity in km / h = total displacement / total time = 8.8 km / 0.5 h = 17.6 km/h

6 0

3 years ago