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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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A 15kg projectile is moving with a velocity of 25m/s calculate its momentum
aliina [53]

The momentum of an object is given by:

p = mv

m is the object's mass and v is its velocity.

Given values:

m = 15kg

v = 25m/s

Plug in the values and solve for p:

p = 15*25

p = 375kg×m/s

8 0
4 years ago
How many nanoseconds does it take light to travel 3.50 ft in vacuum?
Fiesta28 [93]
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
6 0
3 years ago
On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

                                 T  = 2*pi*sqrt(k / m )    ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

                                 F_net = k*x - m*g_x = 0

                                 (k / m) = (g_x / x)    .... 2

- substitute Eq 2 into Eq 1:

                                  2*pi / T = sqrt ( g_x / x )

                                   g_x = (2*pi / T )^2 * x

- Evaluate g_x:

                                  g_x = (2*pi / (19 / 11) )^2 * 0.226

                                  g_x = 3.0 m / s^2

                                 

                       

3 0
3 years ago
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Answer:

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6 0
3 years ago
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Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
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