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2 years ago

8

An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t

he auto's average acceleration?

1 answer:

USPshnik [31]2 years ago

3 0

Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

jekas [21]

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is $w_{h}$ .

We need to calculate the weight of heart of a human

Using given data

$w_{h}=0.5\times w$

Where, h = weight of heart

w = weight of human

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

Hence, The weight of heart of a human is 0.93 lbs.

8 0

2 years ago

An electron is at the origin. (a) Calculate the electric potential VA at point A, x 5 0.250 cm. (b) Calculate the electric poten

saw5 [17]

Answer:

a) V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V c) the sign of the potential change

Explanation:

The electrical potential for a point charge

V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

V_a = -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

V_a = -5.7536 10⁺⁷ V

b) point x = 0.750 cm = 0.750 10-2

Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²

Vb = -1.92 10⁻⁷ V

potemcial difference

ΔV = Vb- Va

V_ba = (-5.7536 + 1.92) 10⁻⁷

V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

ΔV = E d

The force on the particle is

F = q₀ E

F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

V_ba = 0.83 10⁻⁷ V

5 0

2 years ago

A dog travels north for 18 meters, east for 8 meters, South for 27 meters and then west for 8 meters. What is the distance the d

Afina-wow [57]

I'm really not sure if this is right but I'll try.
The distance that the dog traveled is probably all of the distances added up. I would guess that it's 67 meters in total.
The displacement is a little more tricky but you pretty much have to put a mental map in your head. Since East and West are both 8 meters, they cancel each other out. He travels more southern and that means the displacement is 9 meters south of his original location

7 0

2 years ago

A plane took 7 hours to travel 2020 km. For the first 4 hours,

kicyunya [14]

Answer:

-6.6 km/h

Explanation:

In 7hr plane travelled 2020km;

For the first 4hr the average speed was 310km/h;

d=st, s=d/t;

Distance covered in first 4h is d = 310km/h×4h = 1240km;

See the image attached for further solution

8 0

2 years ago

Read 2 more answers