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Sunny_sXe [5.5K]
2 years ago
14

Keisha told her friend that she can get to the town library by walking one mile east, turning left, and then walking a half a mi

le north.
Which best describes how Keisha can improve these instructions to better help her friend find the library?
Physics
2 answers:
Mama L [17]2 years ago
7 0

The best improvement to these instructions that Keisha can make
immediately and easily is to describe where the instructions start from. 

Does the route start from Keisha's house ?  From her friend's house ? 
From my house ? 

Knowledge of the coordinate reference frame makes a big difference,
and it's essential.

nikitadnepr [17]2 years ago
3 0

Answer;

providing a reference point to follow

Explanation;

A reference point is a starting point you choose to describe the location, or position, of an object.

A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. A reference point is some point in space used to describe the position of other things, relative to this reference point.

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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it
vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

F = ma...(I)

Using formula of friction force

F= \mu m g....(II)

Put the value of F in the equation (II) from equation (I)

ma=\mu mg....(III)

a = \mu g

Put the value in the equation (III)

a=0.24\times9.8

a=2.352\ m/s^2

We need to calculate the distance,

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}2.352\times(3.0)^2

s=10.584\ m\ approx\ 11\ m

Hence, The distance is 11 m.

3 0
2 years ago
A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
evablogger [386]

Answer:

\text{heat loss} = 24864.05 \  W/m^2

Explanation:

If

  • T_1, T_2 are temperatures of gasses and liquid in Kelvins,
  • t_1 and t_2 are thicknesses of gas layer and steel slab in meters,
  • h_1, h_2 are convection coefficients gas and liquid in W/m^2 \cdot K,
  • R_c is the contact resistance in m^2 \cdot K/W,
  • and k_1, k_2 are thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

using known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Using the rate equation :

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Similarly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature distribution is shown in the attached image

3 0
2 years ago
The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i
Rainbow [258]

Answer:

4.32\cdot 10^5 hp, 3.22\cdot 10^8 W

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

4F_T - F_d = 0

where

F_T = 322,000 N is the thrust force generated by each engine of the jet

F_d is the drag force

Solving for Fd,

F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N

The velocity of the jet is

v=250 m/s

So, the rate at which the drag force does work (which is the power) is

P=F_d v

and substituting

F_d = 1.288\cdot 10^6 N\\v = 250 m/s

we find

P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W

Converting into horsepower,

P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp

4 0
2 years ago
A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being th
goldfiish [28.3K]

Answer:

7.3 kg m/s

Explanation:

First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:

U=mgh=(0.28 kg)(9.8 m/s^2)(34.3 m)=94.1 J

For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:

K=\frac{1}{2}mv^2=94.1 J

From this equation we can find the velocity of the stone as it leaves your hand:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(94.1 J)}{0.28 kg}}=25.9 m/s

The initial velocity of the stone (before leaving your hand) is zero:

u=0

The impulse received by the stone is equal to its change in momentum, so:

I=\Delta p=m\Delta v=m(v-u)=(0.28 kg)(25.9 m/s-0)=7.3 kg m/s

5 0
2 years ago
For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar
Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

T=\dfrac{30}{2}\ s= 15 \ s

Therefore, total time take to  get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0
2 years ago
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