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stepladder [879]
3 years ago
5

Plz help me fast WITH EXTRA POINTS AFTER SUBMITTING

Physics
1 answer:
bulgar [2K]3 years ago
7 0

Answer:

4 bobux

Explanation:

one bobux

two bobux

three bobux

four bobux

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Scientists use what scale to compare the hardness of minerals?
frozen [14]
Mohs hardness scale. hope this helps
4 0
3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
You are making cookies that call for 3 tablespoons of molasses, but you are having trouble measuring out the thick, syrupy liqui
Andreyy89
You could heat the syrup, grease the spoon with shortening, because if you do, then the molasses won't stick when you pour it out of the spoon into the batter.
3 0
4 years ago
Read 2 more answers
A unit of time sometimes used in microscopic physics is the "shake". One shake equals 10−8 s.
Lady_Fox [76]

Answer:

(a) There are 3.17 times more shakes in a second than seconds in a year.

(b) Humans have existed for 9067.72 universe seconds.

Explanation:

(a) First we calcule how many seconds are in a year. One year have 365 days, one day has 24 hours, one hour has 60 minutes and one minute has 60 seconds, so:

60s*60*24*365=3.1536*10^7s

Now, we calculate how many shakes are in a second:

1s*\frac{1shake}{10^{-8}s}=10^8shakes

Finally, we calculate how many more shakes in a second are there than seconds in a year:

\frac{10^8s}{3.1536*10^7s}=3.17

(b) Using a simple rule of three we can calculate  how many "universe seconds" have humans existed, recall that 1 day has 86400 seconds:

1010 years--------->86400s

106 years----------> x seconds

x=\frac{106y*86400s}{1010y}=9067.72s

3 0
3 years ago
Sobre un gas contenido en un cilindro provisto de un pistón se realiza un trabajo de 7000 Joules, mediante un proceso isotérmico
natita [175]

Answer:

En un proceso isotérmico, es decir, la temperatura no cambia, el trabajo puede escribirse como:

W = n*R*T*Ln(P1/P2)

Donde P1 es la presión inicial y P2 la presión final.

Donde las cantidades:

n =  número de moles

R = constante de los gases ideales

T = temperatura no cambian.

Y sabemos que la ecuación de la energía interna es:

U = C*n*R*T

Donde C es otra constante que depende del gas.

De aca, podemos concluir que ninguna de estas variables cambia en nuestro proceso, por lo que la variación de la energía interna es cero.

U2 - U1 = 0

b) Para el calor cedido o absorbido, la formula básica es:

ΔQ = C*(T2 - T1)

Donde ΔQ es el calor absorbido o cedido por el gas, C es una constante que depende del gas, T2 es la temperatura final del gas y T1 es la temperatura inicial del gas.

Como la temperatura no cambia en el proceso, entonces:

T2 = T1

ΔQ = C*(T2 - T1) = C*0 = 0

No hay calor absorbido ni cedido.

c) Podemos concluir que en un proceso isotérmico la energía interna no cambia, y no hay un intercambio de calor.

8 0
3 years ago
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