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If you separate vector B into its components. How many components will it have? Those components will be called?

Inessa [10]

The vector B will have two components and those components will be called resultant vectors.

<h3>What is a component vector?</h3>

A component vector is a unit vector that represents a given vector in a particular direction.

A vector can be represented in x - direction and y - direction.

x - component of the vector = Bcosθ
y - component of the vector = Bsinθ

Thus, the vector B will have two components and those components will be called resultant vectors.

Learn more about component vectors here: brainly.com/question/13416288

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3 0

2 years ago

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

4 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in-

Leto [7]

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

F = m a₀

a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

x = v₀ t + ½ a t²

x = ½ a₀ t²

t² = 2x / a₀

28² = 2x /a₀ (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

in the direction of axis a, there is a motor so its force is F/2

the acceleration on this axis is

a = F/2m

a = a₀ / 2

so if we use the distance equation

x = v₀ t + ½ a t²

as part of rest v₀ = 0

x = ½ (a₀ / 2) t²

let's clear the time

t² = (2x / a₀) 2

we substitute the let of equation 1

t² = 28² 2

t = 28 √2

t = 39.60 s

4 0

4 years ago

George and Harriot walk with an average velocity of .95 m/s eastward. If it takes them 30

artcher [175]

Answer:

1.71 km

Explanation:

Convert 30 minutes to seconds:

30 min × (60 s / min) = 1800 s

Find the displacement:

0.95 m/s × 1800 s = 1710 m

Convert to kilometers:

1710 m × (1 km / 1000 m) = 1.71 km

5 0

3 years ago

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