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1 year ago

An aircraft on it's take-off run has a steady acceleration of 3m/s^2. How much velocity does it gain 10 seconds?

2 answers:

7 0

To solve this, let's use the formula a=(v''-v')÷(t)

3=(v''-0)÷(10)
3=v÷10

Now isolate for velocity

v"=(3)(10)
v"=30 m/s!

cluponka [151]1 year ago

6 0

Acceleration measures how fast the speed changes over time.

So, over 10s, the aircraft's speed changes by 3 $\frac{m}{ s^{2} }$ * 10s

So, the change is 30 $\frac{m}{s}$

Note that the one of the second units cancels out in the answer.

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The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago

Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?

Oksi-84 [34.3K]

Answer:

$2.32\times 10^{-11}$

Explanation:

First number is $4.48\times 10^{-8}$

Second number is $5.2\times 10^{-4}$

We need to multiply the two numbers.

$4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}$

In scientific notation : $2.32\times 10^{-11}$

Hence, this is the required solution.

8 0

1 year ago

: 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:

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

$F_{No}= 7771.125 \ N$

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

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

$P_d__{D}} = 1828.2 * 79000$

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

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

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

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

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

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

6 0

2 years ago

A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei

QveST [7]

gravitational potential energy is given by formula

$U = mgh$

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

$\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}$

$\frac{U_2}{U_1} =\frac{m_2}{m_1}$

given that

$m_2 = 4 m_1$

now we have

$\frac{U_2}{U_1} = 4$

5 0

2 years ago

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha

Alecsey [184]

Answer:

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0

2 years ago