The answer will b bc grass hopppers cant swin

Suppose a treadmill has an average acceleration of 4.7x10^-3 m/s. a)how much does its speed change after 5min? b)if the treadmil

Mars2501 [29]

Answer:

a)Change in the speed = 1.41 m/s

b)The final speed will be 3.11 m/s

Explanation:

Given that

Acceleration ,a= 4.7 x 10⁻³ m/s²

a)

We know that

v= u + a t

v=final speed ,u=initial speed

t= time ,a= acceleration

Change in the speed

v- u = a t

t= 5 min = 5 x 60 s = 300 s

v- u = 4.7 x 10⁻³ x 5 x 60 m/s

v-u = 1.41 m/s

Change in the speed = 1.41 m/s

b)

Given that

u= 1.7 m/s

v-u = 1.41 m/s

v= 1.7 + 1.41 m/s

v=3.11 m/s

The final speed will be 3.11 m/s

8 0

3 years ago

A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9

ElenaW [278]

Answer:

The convection coefficient is $15456.48\ W/m^{2}K$

Solution:

Mass flow rate, $\dot{m} = 0.4\ kg$

Inner diameter of the tube, d = 0.014 m

Fluid density, $\rho_{f} = 990\ kg/m^{3}$

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, $P_{r} = 8.6$

Heat flux, $\dot{q} = 71,297\ W/m^{2}$

Viscosity, $\mu = 0.00079\ Ns/m^{2}$

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

$A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}$

Determine the velocity of the fluid inside the tube by mass flow rate:

$\dot{m} = \rho_{f}A_{c}v$

$0.4 = 990\times 1.54\time 10^{- 4}v$

v = 2.624 m/s

Determine the Reynold's Number, $R_{e}$ :

$R_{e} = \frac{\rho_{f}dv}{\mu}$

$R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253$

Thus it is clear that $R_{e}$ > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

$N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}$

$N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42$

Also,

$N_{u} = \frac{dh}{K}$

where

h = convection coefficient

Now,

$292.42 = \frac{0.014\times h}{0.74}$

$h = 15456.48\ W/m^{2}K$

7 0

4 years ago

Two objects of equal mass collide on a horizontal frictionless surface. Before the collision, object A is at rest while object B

fenix001 [56]

Answer: 6m/s

Explanation:

Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.

After collision, the two objects will move at the same velocity (v).

Let mA and mB be the mass of the two objects

uA and uB be their velocities before collision.

v be their velocity after collision

Since the two objects has the same mass, mA= mB= m

Also since object A is at rest, its velocity = 0m/s

Velocity of object B = 12m/s

Mathematically,

mAuA + mBuB = (mA+mB )v

m(0) + m(12) = (m+m)v

0+12m = (2m)v

12m = 2mv

12 = 2v

v = 6m/s

Therefore the speed of the composite body (A B) after the collision is 6m/s

7 0

4 years ago

The duckbill platypus and spiny anteater are the only two known

Iteru [2.4K]

Answer:

I believe it is False.

Explanation:

Hope my answer has helped you!

3 0

3 years ago

Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a

DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0

3 years ago