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2 years ago

7

You exert a horizontal force on a heavy piano that rests on a carpeted floor. the piano doesn't budge. if the force you exert on

the piano is the "action", what force is its reaction?

1 answer:

Lerok [7]2 years ago

4 0

The force the piano exerts on you!

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Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created is $P_{avg} = F * v_{avg}$

Explanation:

From the question we are told that

The that the average power is mathematically represented as

$P_{avg} = \frac{W }{\Delta t }$

Where W is is the Workdone which is mathematically represented as

$W = F * s$

Where F is the applies force and s is the displacement due to the force

So

$P_{avg} = \frac{F *s }{\Delta t }$

Now this displacement can be represented mathematically as

$s = v_{avg} * \Delta t$

Where $v_{avg }$ is the average velocity and $\Delta t$ is the time taken

So

$P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }$

=> $P_{avg} = F * v_{avg}$

3 0

2 years ago

29. 2072 Set C Q.No. 10c

Annette [7]

Answer:

$90.2^{\circ}C$

Explanation:

Considering the thermal conductivity of aluminium and brass as $k_{al}=205 W/mK$ and $k_{br}=109 W/mk$ respectively

The temperature at the end of aluminium and brass are given as $T_{al}=150^{\circ}C$ and $T_{br}=20^{\circ}C$ respectively with length of rod L=1.3 m , Length of aluminium $L_{al}=0.8 m$ , length of brass $L_{br}=0.5 m$ and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

$H_{br}=H_{al}$

$k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}$

Upon re-arranging

$T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}$

$(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}$

$T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}$

$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

6 0

2 years ago

A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil

NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ $_{crss}$ = σ $_{Y}$ ( cosФ cosλ) $_{max}$

now by putting values,

= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0

2 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0

2 years ago

At the equator, the earth’s field is essentially horizontal; near the north pole, it is nearly vertical. In between, the angle v

Yanka [14]

Answer:

A current of 0.358A is required

Explanation:

Pls refer to attached document for more details

7 0

2 years ago