What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Can someone help me?!!!!!

Bingel [31]

Answer:

Third Option

$B = -0.5A$

Explanation:

If we have a vector A = ax + by we know that by definition

cA = cax + cby

Where c is a constant.

In this case we have two vectors

$A = 7.6\^x -9.2\^y\\\\B = -3.8\^x + 4.6\^y$

You may notice that vector B has an opposite direction to vector A.

You may also notice that | Ax | is the double of | Bx | and | Ay | is double of |By |

That is to say

$3.8 +3.8 = 7.6\\\\4.6 +4.6 = 9.2$

So the equation that relates to vectors A and B is:

$B = -0.5A$ .

You can verify this relationship by performing the operation

$B = -0.5A$

$-3.8\^x + 4.6\^y = -0.5(7.6\^x -9.2\^y)\\\\\-3.8\^x + 4.6\^y = -3.8\^x + 4.6\^y$

4 0

4 years ago

Read 2 more answers

On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs acco

lianna [129]

Answer:

$\dot n_{f} = 85.177\,rpm$

Explanation:

The expression for the moment of inertia of the person is:

Arms outstretched

$I = \frac{1}{12}\cdot (0.13)\cdot (61\,kg)\cdot (1.40\,m)^{2} + \frac{1}{2}\cdot (0.87)\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 4.546\,kg\cdot m^{2}$

Arms parallel to the trunk

$I = \frac{1}{2}\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 3.736\,kg\cdot m^{2}$

The final angular speed is found by means of the Principle of Angular Momentum Conservation:

$I_{o}\cdot \dot n_{o} = I_{f}\cdot \dot n_{f}$

$\dot n_{f} = \frac{I_{o}}{I_{f}}\cdot \dot n_{o}$

$\dot n_{f} = \left(\frac{4.546\,kg\cdot m^{2}}{3.736\,kg\cdot m^{2}}\right)\cdot (70\,rpm)$

$\dot n_{f} = 85.177\,rpm$

8 0

3 years ago

consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti

ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$$\frac{Energy}{Area} = \sigma\times T^{4}$

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$$E= \sigma\times T^{4}$

Now -

For Star A →

$T_{A}$ = 4900 K

For Star B →

$T_{B}$ = 9900 K

Therefore -

$$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}$

$\frac{T_{B} }{T_{A} }=$ 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }$

E(B) = E(A) x $(\frac{T(B)}{T(A)} )^{4}$

E(B) = E(A) x $2^{4}$

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0

2 years ago

Car A is traveling at a constant speed vA = 130 km / h at a location where the speed limit is 100 km / h. The police offi cer in

kotykmax [81]

Answer:

Distance = 713.303m

Explanation:

To determine the distance required for the police officer to overtake car A, we simply calculate the time t for which the police car and the speeding car have traveled equal distances from point P

5 0

3 years ago

Read 2 more answers

5. Read the final paragraph on the opposite page. What<br> is the speed of the car in km/h?

Vera_Pavlovna [14]

Answer:

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jjjjjtsstujbfryjbctybggn

8 0

3 years ago