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3 years ago

5

Race cars A and B are driving on the same circular racetrack at the same speed of 45.0 m/s. At a given instant car A is on the n

orth side of the track moving eastward and car B is on the south side of the track moving westward. Find the magnitude of the velocity vector of car A relative to car B.

1 answer:

Radda [10]3 years ago

4 0

Answer:90 m/s

Explanation:

Given

Speed of both the car is $v=45\ m/s$

Suppose North side and East side as Positive Y and x axis

so Velocity of car A can be written as

$v_a=45\hat{i}$

Velocity of car B is

$v_b=-45\hat{i}$

Velocity of car A relative to car B is given by

$v_{ab}=v_a-v_b$

$v_{ab}=45\hat{i}-(-45\hat{i})$

$v_{ab}=90\hat{i}$

Magnitude of velocity is

$|v_{ab}|=90\ m/s$

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What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds lat

blagie [28]

The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

The given data in the problem is given by ;

u is the initial speed = 0 m/sec

v is the final speed= 35 m/sec

t is the time interval= 1.2 second

a is the acceleration=? m/sec²

The formula for acceleration is;

$\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2$

Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

To learn more about acceleration, refer to the link;

brainly.com/question/2437624#SPJ2

#SPJ1

5 0

2 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

How many light years away is the sun from the middle of the Millky way

julsineya [31]

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way

4 0

3 years ago

15 points! Answer FAST!

Usimov [2.4K]

Answer:heat brings it up then down

Explanation:

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3 years ago