NEED ANSWER PLEASE!!!!

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi

Aneli [31]

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

$v^2-u^2=2as$

$4^2-5^2=2(a)(3)$

$a=\frac{-3}{2}=-1.5 m/s^2$

after this its final velocity will be zero

$v^2-u^2=2as$

$0^2-4^2=2\times (-1.5)\times s$

s=5.33 m

Total distance=3+5.33=8.33 m

Thus he will not be able to win the game

4 0

2 years ago

A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body, the distance it travelled (is a scalar quantity).

The displacement is the distance in a straight line between the initial and final position (is a vector quantity).

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m). Being its displacement 95 m.

8 0

2 years ago

To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m

hoa [83]

Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

F= ?

W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

3 0

2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i

aniked [119]

Answer:

a. N = 2.49W b. 0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

6 0

2 years ago