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Agata [3.3K]
4 years ago
13

From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco

nd snowball upward with the same initial speed, and this too falls to the ground. neglecting air resistance, the second snowball hits the ground with a kinetic energy of:
a. 100J
b. 200J
c. more than 200J
d. less than 200J
e. none of the above
Physics
1 answer:
Vanyuwa [196]4 years ago
7 0

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

mgh+\dfrac{1}{2}mu^2=mgh'+0          

100=mgh'

At lowest point,

mgh'=K

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).                                                                        

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Joanna learned from customer satisfaction surveys that diners in her restaurant wanted to be able to make substitutions. She has
Dmitry [639]

Answer:

standards

Explanation:

Based on the information provided within the question in regards to the situation at hand it can be said that Joanna is demonstrating a standards gap. this is a gap caused by the difference between the customer service standards a company has created for itself and the expectations the company believes that the customers have for that company. Since Joanna did not tell all the servers of the customers expectations then the ones who do not know will not be able to provide this service to those customers, thus the restaurant will not be able to meet it's customer service standards.

3 0
4 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
kkurt [141]

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

8 0
4 years ago
A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou
tamaranim1 [39]

We have that the amount of the charge q is

q=1.8*10^{-7}

From the Question we are told that

Force F=7.89 x*10^{-7}

Velocity V=2090m/s

Angle \theta=29.4

Magnetic field B=4.23 * 10^{-3} T

Generally, the equation for Force F is mathematically given by

F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}

q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}

q=1.8*10^{-7}

In conclusion

The amount of the charge q is

q=1.8*10^{-7}

For more information on this visit

brainly.com/question/1470439

6 0
3 years ago
Read 2 more answers
List small/average stars<br><br>​
mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0
3 years ago
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