All categories

2 years ago

Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.

Physics

1 answer:

matrenka [14]2 years ago

3 0

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

$s = ut + \frac{1}{2}at^{2}$

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter

You might be interested in

A man pushes a 60.8 kg crate across a rough surface with an applied force of 125 N and at a CONSTANT SPEED.

Sedbober [7]

Answer: Colby we both dumb if we need brainly lol

Explanation:

8 0

3 years ago

Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in

Maksim231197 [3]

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

v = (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

unit (v) = vec (v) / mag (v)

unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

7 0

3 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago

An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this

galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

$\gamma = \frac{Mt}{J}$

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

$J = \frac{1}{2} * m * r^2$

$Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2$

The wheel walls act like annular rings, for these the moment of inertia is:

$J = \frac{1}{2} * m * (re^2 - ri^2)$

$Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2$

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

$J = m * r^2$

$Jtread = 10 * 0.33^2 = 1.09 kg*m^2$

The axle acts like a rod, which is the same as the disk:

$Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2$

The drive shaft acts like a rod too:

$Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2$

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

$\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}$

4 0

3 years ago

PLS ANSWER ASAP

vivado [14]

Answer:

Cart A

Explanation:

Momentum can be computed by finding the product of mass and velocity. To solve this, you can use the formula below to find the greatest momentum:

p = mv

where:

p = momentum (kgm/s) m = mass (kg) v = velocity (m/s)

Because carts are moving along with the weights, we need to consider the whole system. This means that you need to add in the masses and the mass of the cart.

Cart A:

m = 200kg + 0 kg = 200 kg

v = 4.8 m/s

p = 200kg x 4.8 m/s = 960 kg-m/s

Cart B:

m = 200kg + 20 kg = 220 kg

v = 4.0 m/s

p = 220kg x 4.0 m/s = 880 kg-m/s

Cart C:

m = 200kg + 40 kg = 240 kg

v = 3.8 m/s

p = 240kg x 3.8 m/s = 912 kg-m/s

Cart D:

m = 200kg + 60 kg = 260 kg

v = 3.5 m/s

p = 260kg x 3.5 m/s = 910 kg-m/s

As you can see, Cart A has the greatest momentum.

5 0

2 years ago

Read 2 more answers