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Viefleur [7K]
3 years ago
6

1. The maximum running speed (S) in km/hr and the corresponding body mass (m) in kg were

Physics
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

(a) [km/hr/kg]

(b) S[m/s] = (5.7 / 3600) (m[g])

Explanation:

(a) S[km/hr] = 5.7 (m[kg])

Solving for 5.7:

5.7 = S[km/hr] / (m[kg])

5.7 = S/m [km/hr/kg]

(b) S[km/hr] = 5.7 (m[kg])

Convert km/hr to m/s.

S[m/s] = S[km/hr] × [1000 m/km] × [1 hr / 3600 s]

S[m/s] = S[km/hr] / 3.6

3.6 S[m/s] = S[km/hr]

Convert kg to g.

m[g] = m[kg] × [1000 g/kg]

m[g] = 1000 m[kg]

m[g] / 1000 = m[kg]

Substitute.

3.6 S[m/s] = 5.7 (m[g] / 1000)

S[m/s] = (5.7 / 3600) (m[g])

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It takes 185 kj of work to accelerate a car from 23.0 m/s to 28.0 m/s. what is the car's mass?
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6 0
3 years ago
What regions contain the largest concentration of oil wells
lyudmila [28]

Dubai this part of india has a lot of oil is consider one that has more oil wells.

5 0
3 years ago
A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
How much pressure is on the bottom of a pot that holds 20N of soup? the surface area of the pot is 0.05m2
fomenos

Answer:

400

Explanation:

Formula used in solution:

P = \frac{F}{A}

P - pressure,\: F - force,\:  A - area

The given information:

F = 20N

A = 0.05m^{2}

Solution

P = \frac{F}{A} = \frac{20}{0.05} = 20 * 20 = 400

Answer:

Pressure = 400 \: Pascals

8 0
3 years ago
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
3 years ago
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