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Tema [17]
3 years ago
6

a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

ow much momentum does it have when it hits the catcher units
Physics
2 answers:
topjm [15]3 years ago
8 0

Answer:

p = 6.25 kg-m/s

Explanation:

It is given that,

Velocity of the baseball, v = 44.7 m/s

Weight of the baseball, W = 1.4 N

Firstly we need to find the mass of the baseball. It can be calculated as :

m=\dfrac{W}{g}

m=\dfrac{1.4\ N}{9.8\ m/s^2}

m = 0.14 kg

The momentum of an object is equal to the product of mass and velocity. It is given by :

p=m\times v

p=0.14\ kg\times 44.7\ m/s

p = 6.25 kg-m/s

So, the momentum it have when it hits the catcher is 6.25 kg-m/s. Hence, this is the required solution.

kumpel [21]3 years ago
3 0
1.4 N is a weight so calculating it's mass
1.4/9.8 = 0.1428 kg
momentum will be 0.1428*44.7 = 6.38 kgm/s
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A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val
Lostsunrise [7]

Answer:

1.25\ \mu\text{g/m}^3

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}

\dot{m} = Mass flow rate of pollutant = 25 g/s = 25\times 10^6\ \mu\text{g/s}

Concentration is given by

C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3

The steady state concentration of pollutants in the valley, is 1.25\ \mu\text{g/m}^3.

6 0
3 years ago
How can a calculated height be greater than an actual height?
sammy [17]

Answer:

mesuring heigh and weight is important

3 0
2 years ago
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
vagabundo [1.1K]

Answer:

R_total = 14.57 Ω ,  V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance  

         R_{eq} = ∑ R_{i}

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

         R_B = 20 Ω

Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

          1 / R_BC = 1/20 + 1/15 = 0.116666

          R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

          R_total = R_A + R_BC

          R_total = 6 + 8.57

          R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

           10 = V_a + V_BC

           10 = i R_a + i R_BC = i (R_a + R_BC)

           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

           i = 0.6863 A

The current in the series circuit is constant

          V_BC = i R_BC

          V_BC = 0.6863 8.57

          V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

     Branch C

             V_BC = i R_C5

             i = V_BC / R_C5

             i = 5.8819 / 15

             i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

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Plssss help I need good grade!!!! Plssss and thank u
shepuryov [24]

Answer:

the answer is 'b'

According to me

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