All categories

2 years ago

I need help with number 7 and 8. Gravity is 10N NOT 9.8N for these problems

Physics

1 answer:

skad [1K]2 years ago

4 0

What do you need to find?The problem is not clear?!

You might be interested in

The graph illustrates the activity level of three common digestive enzymes, across a range of pH values. Which enzyme is likely

vaieri [72.5K]

Answer:

(A) Pepsin

Explanation:

From the graph it is clear that pepsin is the only enzyme which works in highly acidic condintion in the digestive system.

less than 7 the liquid is acidic
above 7 the liquid is basic
at 7 the liquid is neutral

It has an optimum pH of about 1.5 at which its activity level is 8.5 as shown in graph.

7 0

3 years ago

Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic

marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = $1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 0.063 m³/s

flow rate in liters per minute

1 gallon = 3.78541 L

Q = $1000\times \dfrac{3.78541\ m^3}{1\ gallon}$

Q = 3785.41 m³/min

flow rate in cubic feet per second

1 gallon = 0.133681 ft³

1 min = 60 s

Q = $1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 2.23 ft³/s

4 0

2 years ago

Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto

Irina-Kira [14]

Answer:

W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

V = A L

V = π (R_ext ² - R_int ²) L

let's calculate

V = π (4² - 2²) 10⁻⁴ 3

V = 1.13 10⁻² m³

m = ρ V

m = 8966 1.13 10⁻²

m = 1.01 10² kg

the weight of the tube

W = mg

W = 1.01 10² 9.8

W = 9.93 10² N

4 0

2 years ago

The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False

madreJ [45]

True

In fact, the weight of an object on the surface of the Earth is given by:
$F=mg$
where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
$F=mg=(3.0 kg)(9.81 m/s^2)=29 N$

8 0

2 years ago

Escribe un listado de 5 acciones que promueven la interculturalidad y la inclusión

allochka39001 [22]

de mategoloalterfsqol

6 0

2 years ago