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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 rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump
ruslelena [56]

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

To calculate efficiency we need to find power output of electric pump.

We can use,

Work done = Energy change

Work done per second = Energy change per second

Work done per second = Power

Therefore, Power = Energy change per second

                              = Change in potential energy of water per second

                              =mgh / t

                              = 200× 10×6 / 10

                              = 1200 W = 1.2 kW

Now use the first equation to find efficiency,

Efficiency = \frac{1.2}{1.5} × 100%

                = 80 %

8 0
3 years ago
Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3​
kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0
3 years ago
A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor
Serggg [28]

Answer:

a = 1 m/s²  and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

      Ty - W = 0        

      Ty = w

X axis

     Tx = m a

With trigonometry we find the components of tension

    Sin θ = Ty / T

    Ty = T sin θ

    Cos θ = Tx / T

    Tx = T cos θ

We calculate the acceleration with kinematics

   Vf = Vo + a t

   a = (Vf -Vo) / t

   a = (20 -10) / 10

   a = 1 m/s²

We substitute in Newton's equations

     

  T Sin θ = mg

  T cos θ = ma

We divide the two equations

  Tan θ = g / a

  θ = tan⁻¹ (g / a)

  θ = tan⁻¹ (9.8 / 1)

  θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0
2 years ago
A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no
ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, \vec{F_{net}} = \vec{F'} - \vec{F}

\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

\vec{F_{net}} = 7 N towards North

4 0
3 years ago
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