My grandmother lives 100m north of me. If I get in my car and drive to her house in 4s,

What what is the direction of an object’s acceleration if it is being pulled upward by a rope at a constant speed?

Arada [10]

Answer:

the direction is towards upward.i think.

7 0

3 years ago

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the

Eduardwww [97]

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0

3 years ago

-What can you say about the snowboarder’s kinetic energy as he moves?

Damm [24]

Answer:

His kinetic energy increases, potential energy decreases

The sum of kinetic and potential energy is a constant at any instant before he comes to rest.

Explanation:

Snowboarder is starting from a height and moving to the down direction. As he moves down his velocity increases, we know that kinetic energy is given by the expression $\frac{1}{2} mv^2$ , so as he moves his kinetic energy increases.

When the snowboarder is starting his potential energy is maximum(Potential energy = mgh), as he comes down his potential energy decreases.

Based on this we can conclude that the sum of potential energy and kinetic energy is a constant at any instant for a snowboarder before he comes to rest.

mgh+ $\frac{1}{2} mv^2$ = Constant

7 0

4 years ago

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

Lena [83]

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$ , $2$\hbar$ , $3$\hbar$$ , etc. Bosons particles always have asymmetric wave function and there is exchange of particles.

1) It does not obey Fermi_ Dirac statistics

2) It obeys Bose-Einstein statistics

3) The object can have intrinsic spin $2$\hbar$

4) Yes the Bosons particle is always symmetric with exchange of particles

5) No Bosons particle are symmetric and not asymmetric

5 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago