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3 years ago

Can someone please help me with this problem

Physics

1 answer:

sukhopar [10]3 years ago

8 0

Answer:

resultant is equal to the sum of A vector or B vector and draw resultant in order that the tail of resultant vector concides with tail of vector a and head of resultant concides with the head of vector b

Explanation:

You might be interested in

Which statement about the effects of medium on the speed of a mechanical wave is true?

Anastasy [175]

C. A mechanical wave generally travels faster in gases than liquids.

7 0

3 years ago

Suppose the activity of a sample of radioactive material was 100Bq at the start. What would you divide 100Bq by to obtain the ac

myrzilka [38]

If n=1 you divide by 2
If n=2 you divide by 4
If n=3 you divide by 8
so any n you divide by 2 to the power n

4 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

Two samples of water are mixed together.

horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x) Remove the brackets.

0.5x = 150 - 1.5x Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x Combine like terms

2x = 150 Divide by 2

x = 75

Answer

6 0

3 years ago

A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer

pogonyaev

Answer: -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

Then the acceleration equation of the ferry will be:

a(t) = -A

(the negative sign is because this acceleration is in the opposite direction with respect to the movement of the ferry)

To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

v(t) = -A*t + v0

where v0 is the initial velocity of the ferry, v0 = 12m/s.

v(t) = -A*t + 12m/s

For the position equation of the ferry we need to integrate again over time:

p(t) = (-A/2)*t^2 + (12m/s)*t + p0

Where p0 is the initial position of the ferry, in this case, it can be zero, because it will depend on where we put the origin on our coordinate axis.

then p0 = 0m

P(t) = (-A/2)*t^2 + (12m/s)*t

The ferry will come to rest at the moment when it's velocity is equal to zero, this will happen when:

v(t) = 0m/s = -A*t + 12m/s

We need to find the value of t.

A*t = 12m/s

t = (12m/s)/A

Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p( (12m/s)/A) = 2500m = (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

2500m*A = (72 m^2/s^2)

A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

-0.0288 m/s^2

4 0

3 years ago