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ExtremeBDS [4]
3 years ago
9

Just before the ball leaves her hand, what is its centripetal acceleration?

Physics
1 answer:
erica [24]3 years ago
5 0
A = \frac{ v^{2} }{r} = ω²r
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You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you
Leona [35]
Http://www.calculator.net/pace-calculator.html?ctype=distance&ctime=05%3A00%3A00&cdistance=5&cdistanceunit=Miles&cpace=02%3A00%3A00&cpaceunit=tpm&printit=0&x=87&y=24 a pace calculator
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3 years ago
A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

8 0
3 years ago
A materials density is the same , no matter how large or small the sample is or what it’s shape is as long as it is solid unifor
Scorpion4ik [409]

Answer:

See the explanation below

Explanation:

Density is defined as the relationship between mass and volume, i.e. the following equation can be used:

density = m/v

where:

density [kg/m^3]

m = mass [kg]

v = volume [m^3]

If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.

m = density*v

To understand this concept more clearly, let's use the following example:

We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.

1000 = m /1

m = 1000*1 = 1000 [kg]

Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.

1000 = 500/v

v = 500/1000

v = 0.5 [m^3]

We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.

8 0
3 years ago
You decide to ride your bike for 30 minutes, raising your heart rate to the middle of your target heart rate zone. What type of
a_sh-v [17]
It would have to be cardio i think not totally sure
4 0
4 years ago
Read 2 more answers
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

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

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

6 0
3 years ago
Read 2 more answers
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