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mariarad [96]
3 years ago
5

An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that

Physics
2 answers:
SIZIF [17.4K]3 years ago
6 0

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

nevsk [136]3 years ago
6 0

Answer:

14 s

Explanation:

Given:

v₀ = 0 m/s

a = 34 m/s²

Δx = 3400 m

Find: t

Δx = v₀ t + ½ at²

(3400 m) = (0 m/s) t + ½ (34 m/s²) t²

t = 14.1 s

Rounded to two significant figures, it takes 14 seconds.

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Suppose for your cookout you need to make 100 hamburgers. You know that 2.00 pounds will make 9 hamburgers. How many pounds will
VMariaS [17]
2 pounds = 9 burgers figure out ow many 9's you can get out of 100: 100/9=11 but that only makes 99 you need 100 so we would add another one making 12. now multiply 12 by 2: 12·2=24. You would need 24 pounds of meet :)

8 0
2 years ago
I need help with this problem :(​
Reil [10]

Answer:

it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.

7 0
2 years ago
On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
vichka [17]

Answer:

326149.2 KJ

Explanation:

The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:

Q = m*cv*ΔT

Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.

A 65000 L swimming pool will have a mass of:

65000L *\frac{1m^3}{1000L} * \frac{1000kg}{1m^3} = 65000 kg

The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.

We replace the data and get:

Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ

3 0
3 years ago
The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th
bagirrra123 [75]

The velocity of the wave on the string is given by

v=\sqrt{\frac{T}{\frac{m}{L}}}  \\  v=\sqrt{\frac{TL}{m}}

Solving the above equation,

v^2=\frac{TL}{m} \\  L=\frac{v^2m}{T}

The frequency of the wave f=300 and wave length is 0.76

The velocity is v=(300)(0.76)=228

Substituting numerical values,

L=\frac{228^2(0.0059)}{200}\\ T=1.534

The length of the string is 1.534 m

4 0
3 years ago
Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =
Ann [662]

Answer:

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0
3 years ago
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