An object is lifted upward at a constant speed. The gain in potential energy of the object is due to the

On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

Zanzabum

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

4 0

3 years ago

Arrange the following decisions accordingly. Be guided by the chronological circumstances happened from the start of the game. U

Snezhnost [94]

Answer: Use Roman Numerals in answering. 11. - 27310189. ... Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo ...

Explanation:. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag

3 0

2 years ago

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter su

balandron [24]

The new period will be 2.486 days.

<h3>What is the period?</h3>

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

$\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days$

Hence, the new period will be 2.486 days.

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ1

8 0

1 year ago

suppose the same chest is at rest. you push it horizontally with force of 110N but it does not budge. What is the contact force

Radda [10]

Answer:

110 N

Explanation:

When a force is applied on a body and body does not move, it means the body remains at rest.

In this condition, there is a contact force between the body and the floor which is called static friction.

Th static friction force is a self adjusting force and comes into play when the body is at rest.

Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.

So, the contact force between the chest and the floor is 100 N.

4 0

3 years ago

Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

5 0

2 years ago