A box is pushed horizontally with constant speed across a rough horizontal surface.

The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. In this problem you can appro

Pavel [41]

Answer:

A) $A=12.2480\ m^2$

B) $12.2480\pm 0.1029\ m^2$

Explanation:

<u>Given:</u>

Length of the room $l= 3.92 ± 0.0035$

Width of the room $w= 3.15 ± 0.0055$

A) Let A be the area of the room

$A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2$

B)We will calculate uncertainty in each dimension

%uncertainty in length $=\dfrac{0.0035}{3.92}\times 100=0.0892\ %$

%uncertainty in width = $\dfrac{0.0055}{3.15}\times 100=0.0174%$

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area= %uncertainty in length + %uncertainty in width

%uncertainty in Area $=0.0892\ % + 0.0174\ %$

%uncertainty in Area=0.0106

Uncertainty in Area $=0.0106\times 12.2480=0.1029\ \rm m^2$

There Area is $12.2480 ± 0.1029\ \rm m^2$

7 0

2 years ago

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

So

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

So

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

3 0

2 years ago

Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa, 518 oC, and mass flow rate of 5,322 kg/h and exits the system at 96

Orlov [11]

Answer:

The velocity at the nozzle at inlet $V_{1}$ = 3584 $\frac{m}{sec}$

Explanation:

Pressure at inlet $P_{1}$ = 1 × $10^{6}$ Pa

Temperature at inlet $T_{1}$ = 518 ° c = 791 K

Mass flow rate = $\frac{5322}{60}$ $\frac{kg}{sec}$ = 88.7

Gas constant for carbon die oxide is R = 189 $\frac{J}{kg k}$

Mass flow rate inside the nozzle is given by the formula = $\frac{P_{1} }{R T_{1} }$ × $A_{1}$ × $V_{1}$

⇒ $P_{1}$ = = 1 × $10^{6}$ Pa

⇒ R $T_{1}$ = 791 × 189 = 149499 $\frac{J}{kg}$

⇒ $A_{1}$ = 0.0037 $m^{2}$

Put all the above values in above formula we get,

⇒ 88.7 = $\frac{10^{6} }{149499}$ × 0.0037 × $V_{1}$

⇒ $V_{1}$ = 3584 $\frac{m}{sec}$

This is the velocity at the nozzle at inlet.

3 0

2 years ago

Quartz has an index of refraction of 1.46. Diamond has in index of refraction of 2.42. In which material does a light ray enteri

Ksivusya [100]

<span>Diamond slowdown light more than Quartz , because diamonds have a greater index of refraction. Light will bend when its move from one medium to another. The Index of Refraction of Material is found by comparing the speed of light in their respective mediums.</span>

4 0

2 years ago

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Two states of matter are described below. State A: Cannot be compressed and retains its shape State B: Highly compressible Which

telo118 [61]

Answer:

State A = piece of metal; State B = air

Explanation:

For the three main states of matter here's how it breaks down.

Solid - Cannot be compressed and retains its shape

Liquid - Cannot be compressed and does not retain its shape

Gas - Compressible and does not retain its shape.

Knowing this State A has to be solid. Only one of the options has A as a solid, so that's the answer. Worth knowing state B is a gas though, only one compressible, just like solid is the only one that retains its shape.

7 0

3 years ago

Read 2 more answers