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Lorico [155]
3 years ago
7

HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Diano4ka-milaya [45]3 years ago
6 0
B, it has a drop, so it is going to increase in speed
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Need help today please​
Eddi Din [679]

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

3 0
3 years ago
A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the total distance walked by the hiker?
AnnyKZ [126]

Answer:

22 km

Explanation:

11km + 11km= 22km

3 0
3 years ago
Read 2 more answers
You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
A ball of mass m and weight mg is held at rest by two strings. One string makes an angle θ with the vertical. The other is horiz
malfutka [58]

Answer:

W = M g          weight of ball

T cos θ  = W         balancing vertical forces

T sin θ = F            balancing horizontal forces

tan θ = F / W        dividing equations

F = W tan θ          when θ equals zero F equals zero

6 0
1 year ago
Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a
DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0
3 years ago
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