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Natasha_Volkova [10]
3 years ago
13

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Physics
1 answer:
VMariaS [17]3 years ago
8 0

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

v=\dfrac{d}{t}

Where. d = distance

t = time

Put the value into the formula

v=\dfrac{18.288}{2}

v=10\ m/sduring that run

(b). We need to calculate their velocity

Using formula of speed

v=\dfrac{\Delta d}{\Delta t}

Put the value into the formula

v=\dfrac{22.86-36.58}{2}

v=-6.86\ m/s

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

d=vt

Put the value into the formula

d=6.86\times1.3

d=8.91\ m

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

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Torque can cause the angular momentum vector to rotate in UCM. This motion is called ___________.
emmainna [20.7K]

Torque can cause the angular momentum vector to rotate in UCM. This motion is called _Conservation of Angular momentum__________.

Answer:

Conservation of Angular momentum

Explanation:

The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly changing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant.

The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

Key Points

When an object is spinning in a closed system and no external torques are applied to it, it will have no change in angular momentum.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation.

If the net torque is zero, then angular momentum is constant or conserved.

Angular Momentum

The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero. We can see this by considering Newton’s 2nd law for rotational motion:

τ→=dL→dt, where  

τ is the torque. For the situation in which the net torque is zero,  

dL→dt=0.

If the change in angular momentum ΔL is zero, then the angular momentum is constant; therefore,

⇒

L  =constant

L=constant (when net τ=0).

This is an expression for the law of conservation of angular momentum.

Example and Implications

An example of conservation of angular momentum is seen in an ice skater executing a spin,  The net torque on her is very close to zero,

because (1) there is relatively little friction between her skates and the ice, and (2) the friction is exerted very close to the pivot point.

Conservation of angular momentum is one of the key conservation laws in physics, along with the conservation laws for energy and (linear) momentum. These laws are applicable even in microscopic domains where quantum mechanics governs; they exist due to inherent symmetries present in nature.

7 0
3 years ago
Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (
Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud.  The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

    2. A galaxy collision results tostripping of gas.

    3. The protogalactic cloud rotates in a very slow motion.

    5. The density of protogalactic cloud is very high.

    6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

    1. The protogalactic cloud has high angular momentum.

    4. Most of the protogalactic gases settles down into a disk.

6 0
3 years ago
weegy a 7.5kg block is placed on a table. if its bottom surface area is 0.6m2 , how much pressure does the block exert on the ta
Lesechka [4]

The pressure exerted by the block on the table is given by:

p=\frac{W}{A}

where W is the weight of the box, and A is the bottom surface area of the box.

The weight of the box is: W=mg=(7.5 kg)(9.81 m/s^2)=73.6 N

Substituting into the first equation, we find the pressure:

p=\frac{W}{A}=\frac{73.6 N}{0.6 m^2}=122.7 Pa

4 0
4 years ago
The potential energy of a 35 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
Anettt [7]

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,  

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters  high.

6 0
3 years ago
SCIENCE HELP MEEEE PLEASE
maks197457 [2]

Answer:

lol no

Explanation:

8 0
3 years ago
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