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san4es73 [151]
3 years ago
5

Need help today please​

Physics
1 answer:
Eddi Din [679]3 years ago
3 0

6) True: the car is exerting an equal and opposite force on the truck

7) True: the astronaut can drift back by throwing the tool forward

8) True: the bug exerts an equal and opposite force on you

9) Second Law

Explanation:

6)

This problem can be solved by applying Newton's third law of motion, which states that:

"When an object A exerts a force (action) on another object B, then object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the car and the truck as object A and object B. Here we are told that the truck is exerting a force on the car: therefore, according to Newton's third law, the car is also exerting an equal and opposite force on the truck.

Therefore, the statement is true.

7)

This problem can be also solved by thinking in terms of the Newton's third law of motion.

In fact, at the beginning the astronaut is drifting away from the space station. When he throws away the tool, in the forward direction (away from the space station), he exerts a force on the tool: we can identify this force as the action force, and its direction is away from the space station.

As a result, according to Newton's third law, the tool will also exert a reaction force (equal and opposite) on the astronaut: therefore, the force exerted by the tool on the astronaut is toward the space station, and therefore the astronaut will be pushed back towards the station.

Therefore, the statement is true.

8)

Again, this problem can also be explained using Newton's third law.

In fact, the moment you step on the bug, your foot exerts a force (the action force) on the bug, pushing downward.

As a result, according to Newton's third law, the bug exerts back on you an equal and opposite force (upward). The reason you don't feel this force at all is that your mass is much larger than that of the bug, therefore your acceleration is negligible.

Therefore, the statement is true.

9)

This problem can be explained by using Newton's second law, which states that the net force acting on an object is equal to the product between its mass and its acceleration. Mathematically:

F=ma

where

F is the net force

m is the mass

a is the acceleration

In this problem, the ball slows down as it crosses the field: this means that it has an acceleration (more precisely, a negative acceleration). According to the law, this means also that there is a net, unbalanced force acting on it, in the direction opposite to the motion of the ball. In fact, this force is the force of friction between the ball and the surface.

Learn more about Newton laws of motion:

brainly.com/question/11411375

brainly.com/question/3820012

#LearnwithBrainly

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

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3 0
3 years ago
it took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 miles per hour. How many miles lie
Art [367]

Given:

Time: 3.5 hrs

Velocity: 120 miles/hr

Now Distance=  Speed × Time

Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.

Hence Distance = Velocity x time

Distance = 3.5 × 120 = 420 miles

7 0
3 years ago
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
Not a question, but I need help on the last two questions I posted, preferably no links, please...
Andrei [34K]

Answer:

ok...

Explanation:

7 0
3 years ago
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