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3 years ago

Please PLEASE HELP ASAP

Physics

1 answer:

Studentka2010 [4]3 years ago

3 0

Answer:

D) The tray would need to be raised higher before the items start to slide.

Explanation:

The tray would need to be raised higher due to the gritty sandpaper causing it to have more grip or fraction between the two surfaces. In the first experiment it would have a smooth surface to slide across but in the second experiment it would have a rougher surface to slide across.

Example: It would be like taking two different kinds of shoes like tennis shoes and football cleets. Then seeing which one has more traction when going up a hill.

Hope this helps!!

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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

Compare and Contrast Potential Energy with Thermal Energy

Brilliant_brown [7]

Potential energy is energy stored due to its position. Thermal energy is energy released as heat

8 0

3 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

7 0

3 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

3 0

3 years ago

I need help with questions 6-8. Thank you!! Image is attached

Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

In a displacement-time plot, the slope of the line is given by

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the change in the y-variable, so it is the displacement

$\Delta x$ is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

$v=m=\frac{d}{t}$

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

$v=\frac{0-(-10)}{12-8}\sim 2.5 m/s$

So the closest option is b) 2.7 m/s.

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

In the graph, the overall displacement of each object is given by the change in the y-variable, $\Delta y$ .

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y, from -15 m to 30 m, so

$\Delta y=30-(-15)=45 m$

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

$|\Delta y| = 25 -(-15)=40 m$

Finally, object C has displacement

$\Delta y = 20-(-5)=25 m$

While object A has displacement zero. Therefore, the correct option is

b) B

3 0

3 years ago