A 47.3 ml sample of alcohol has a mass of 37.32 g. what is its density in g/ml?

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

so

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

2 years ago

Newton’s Law of Cooling. Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportiona

nexus9112 [7]

Answer:

After 30 minutes, the temperature of the body is: T₁₀ = 311.60 K

After 60 minutes, the temperature of the body is: T₂₀ = 298.18 K

Explanation:

Given that:

$\dfrac{dT}{dt}= K \bigg [ M(t) -T(t)\bigg]$

where;

K = 0.04

M(t) = 293

Then;

$\dfrac{dT}{dt}= 0.04 \bigg [ 293 -T\bigg]$

$\dfrac{dT}{dt}= 11.72 -0.04 \ T$

Using Euler's Formula;

$T_{n+1} = T_n + hf( t_n, T_n)$

where;

$f(t_n,T_n) = 11.72 - 0.04 T_n$

Then;

$T_{n+1} = T_n + 3.0 (11.72-0.04 \ T_n)$

$T_{n+1} = 0.88T_n + 35.16 \ ---(1)$

At initial state $t_0$ (0); $T_0$ = 360

At t₁ = 3.0 when T₀ = 360

$T_1= 0.88 T_o + 35.16$

$T_1= 0.88 (360) + 35.16$

$T_1 = 351.96 \ K$

At t₂ = 6.0 when T₂ = 0.88T₁ + 35.16

T₂ = 0.88(351.96) + 35.16

T₂ = 344.89 K

At t₃ = 9.0 when T₃ = 0.88T₂ + 35.16

T₃ = 0.88(344.89) + 35.16

T₃ =338.66 K

At t₄ = 12.0 when T₄ = 0.88T₃ + 35.16

T₄ = 0.88(338.66) + 35.16

T₄ = 333.18 K

At t₅ = 15.0 when T₅ = 0.88T₄ + 35.16

T₅ = 0.88(333.18) + 35.16

T₅ = 328.36 K

At t₆ = 18.0 when T₆ = 0.88T₅ + 35.16

T₆ = 0.88(328.36) + 35.16

T₆ = 324.12 K

At t₇ = 21.0 when T₇ = 0.88T₆ + 35.16

T₇ = 0.88(324.12) + 35.16

T₇ = 320.39 K

At t₈ = 24.0 when T₈ = 0.88T₇ + 35.16

T₈ = 0.88(320.29) + 35.16

T₈ = 317.02 K

At t₉ = 27.0 when T₉ = 0.88T₈ + 35.16

T₉ = 0.88(317.02) + 35.16

T₉ = 314.14 K

At t₁₀ = 30 when T₁₀ = 0.88T₉ + 35.16

T₁₀ = 0.88(314.14) + 35.16

T₁₀ = 311.60 K

At t₁₁ = 33.0 when T₁₁ = 0.88T₁₀ + 35.16

T₁₁ = 0.88(311.60) + 35.16

T₁₁ = 309.37 K

At t₁₂ = 36.0 when T₁₂ = 0.88T₁₁ + 35.16

T₁₂ = 0.88(309.37)+ 35.16

T₁₂ = 307.41 K

At t₁₃ = 39.0 when T₁₃ = 0.88T₁₂ + 35.16

T₁₃ = 0.88( 307.41) + 35.16

T₁₃ = 305.68 K

At t₁₄ = 42.0 when T₁₄ = 0.88T₁₃ + 35.16

T₁₄ = 0.88(305.68) + 35.16

T₁₄ = 304.16 K

At t₁₅ = 45.0 when T₁₅ = 0.88T₁₄ + 35.16

T₁₅ = 0.88(304.16) + 35.16

T₁₅ = 302.82 K

At t₁₆ = 48.0 when T₁₆ = 0.88T₁₅ + 35.16

T₁₆ = 0.88(302.82) + 35.16

T₁₆ = 301.64 K

At t₁₇ = 51.0 when T₁₇ = 0.88T₁₆ + 35.16

T₁₇ = 0.88(301.64) + 35.16

T₁₇ = 300.60 K

At t₁₈ = 54.0 when T₁₈ = 0.88T₁₇ + 35.16

T₁₈ = 0.88(300.60) + 35.16

T₁₈ = 299.69 K

At t₁₉ = 57.0 when T₁₉ = 0.88T₁₈ + 35.16

T₁₉ = 0.88(299.69) + 35.16

T₁₉ = 298.89 K

At t₂₀ = 60 when T₂₀ = 0.88T₁₉ + 35.16

T₂₀ = 0.88(298.89) + 35.16

T₂₀ = 298.18 K

4 0

2 years ago

For a given force as time ___

myrzilka [38]

Answer:

Increases

Explanation:

For a given force, as the time increases, the impulse related to that force increases also.

Impulse and time are directly related to one another. Impulse is mathematically given as:

I = F t

I is the impulse

F is the force

t is the time taken

So, as both force and time increases, the value of the impulse must increase.

8 0

2 years ago

Which of these properties change if the size of an object changes?

german

The correct answer is
<span>Density

In fact, the density of an object is the ratio between its mass m and its volume V:
</span> $d= \frac{m}{V}$
<span>if the size of the object changes, its volume V changes as well, therefore the density d must change.</span>

7 0

2 years ago

A new company is making security cameras. These security cameras will be different than other security cameras because instead o

babunello [35]

Answer:

b

Explanation:

4 0

1 year ago