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3 years ago

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Calculate the horizontal force that must be applied to a 1,300 kg vehicle to give it an acceleration of 2.6 m/s² on a level road

.
500 N

3,380 N

1,302.6 N

0.002 N

1 answer:

ANEK [815]3 years ago

4 0

The answer is 3,380 N

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A 5000kg car traveling at 40m/s crashes into a wall and comes to a complete stop in 10s. What was the force on the wall? solve s

IrinaK [193]

Answer:

20,000 N

Explanation:

First find the acceleration:

a = Δv / Δt

a = (0 − 40 m/s) / 10 s

a = -4 m/s²

Next use Newton's second law to find the force on the car:

F = ma

F = (5000 kg) (-4 m/s²)

F = -20,000 N

According to Newton's third law, the force on the wall is equal and opposite the force on the car.

F = 20,000 N

4 0

3 years ago

An object has a constant velocity of 20 m/s to the north, and there are two forces acting on it. What is this situation called

Llana [10]

When an object move with a constant velocity and there are forces acting on it, this is called unbalanced force. The forces acting on an object is a net force, which is capable of changing speed and/or direction of motion of an object. However, the two forces acting on the object can be added together to give the resultant force, which is a single force that has the same effect on the object as all the individual forces acting together.

5 0

3 years ago

Read 2 more answers

Steam enters a long, horizontal pipe with an inlet diameter of D1 = 16 cm at 2 MPa and 300°C with a velocity of 2.5 m/s. Farther

polet [3.4K]

Answer:

m = 0.4005 kg/s

Q_out = 45.1 KJ/s

Explanation:

Given

Pipe inlet diameter D1 = 16 cm

Steam inlet pressure P1 = 2 Mpa

Steam inlet temperature T1 = 300 °C

Pipe outlet diameter D2 = 14 cm

Steam inlet velocity V1 = 2.5 m/s

Steam outlet pressure P2 = 1.8 MPa

Steam outlet temperature T2 = 250 °C

Required

Determine

(a) The mass flow rate of steam.

(b) The rate of heat transfer.

Assumptions

Kinetic and potential energy changes are negligible.

This is a steady flow process.

There is no work interaction.

Solution

Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C

vl = 0.12551 m^3/Kg

h1 = 3024.2 KJ/Kg

The mass flow rate of steam could be defined as the following

m = 1/v1*A1*V1

m = 0.4005 kg/s

Part b We take the pipe as our system.The energy balance could be defined as the following

E_in -E_out =ΔE_sys = 0

E_in = E_out

mh_1 = Q_out + mh_2

Q_out = m(h_1-h_2)

From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C h2= 2911.7 KJ/Kg The heat transfer could be defined as the following

Q_out = m(h_1-h_2)

Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s

6 0

3 years ago

(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c

andreyandreev [35.5K]

Answer:

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

Explanation:

The formula for the surface of the circle is:

$A(r) = \pi\cdot r^{2}$

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$

Finally, variables are replaced by known data:

$\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )$

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

4 0

4 years ago

Read 2 more answers

Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

$E_1_x = 0V/m\\E_1_y=100V/m$

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

$E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m$

∑x: $E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m$

∑y: $E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m$

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

$||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees$

or 27.23° to the left of E1.

8 0

3 years ago