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Lena [83]
3 years ago
13

1.a baseball hits a home run, according to Newton's laws of motion which of the following statements could be true in this scena

rio:
statement 1: the bat pushes on the ball with the same force as the ball pushes on the bat.

statement 2: Newton's 2nd law explains why the ball travels farther than the bat.

statement 3: the bat hit the ball causing the ball to change direction. this is an example of unbalanced force in Newtons first law of motion.

choose all statements that are true.

2. A bowling ball on earth weighs 11 pounds and has a mass of 5kg. The gravitational pull on the moon is 1/6 of the gravitational force on earth. what would the mass of the bowling ball be on the moon?

3.how much force would you need to use in order to accelerate an 80kg shopping cart to 0.2 m/s^2 (f=ma)?
Physics
1 answer:
Fittoniya [83]3 years ago
8 0
<h2>Answer to Q1:  </h2>

<u>The correct option is </u><u>bat pushes on the ball with the same force as the ball pushes on the bat. </u>

<h2>Explanation: </h2>

According to Newton’s law action and reaction are equal in magnitude but opposite in direction so the same case will apply here. The bat pushes on the ball with the same force as the ball pushes on the bat just the direction is opposite.

<h2>Answer to Q2 </h2>

Mass will remain the same

<h2>Explanation: </h2>

The gravitational pull is related to the weight of objects. Weight is the force with which earth attracts everything towards its centre. And mass is the quantity of matter inside the body. So on the moon the quantity of matter inside a body does not change therefore the mass would remain the same.

<h2>Answer to Q3 </h2>

Force required is 16 Newton

<h2>Explanation: </h2>

Mass= 80 kg

Acceleration = a = 0.2

So F = ma

Putting the values

F= 80 x 0.2

F = 16 N


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3 years ago
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When two oceanic plates converge, the plate with a greater density will subduct. This plate will partially melt, causing the for
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Island arc

Explanation:

When two oceanic plates share a convergent type of plate boundary, the denser oceanic plate will subduct below the less dense oceanic plate. This will result in the formation of the subduction zone, where the rocks are being pulled down to the mantle. This subduction zone is typically marked by the presence of a narrow depression commonly known as an oceanic trench, that lies just above the zone.

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3 0
3 years ago
An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

                                                                     = 780/130

                                                                     = 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

                                                                                          = 910 J

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5 0
2 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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Answer:

Yes, they are.

Explanation:

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