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3 years ago

An object that is moving must have a change in A) its speed. B) its position. C) its acceleration. D) its applied force.

1 answer:

8 0

An object that's moving doesn't necessarily change its speed or acceleration. Also, the force applied to it doesn't need to change ... in fact, a moving object doesn't need ANY force applied to it in order to keep moving.

But any moving object WILL have a change in its position ... THAT's how you know it's moving, and that's WHY you say "It's moving !". (choice-B)

You might be interested in

How can you measure the distance an object has moved?

Naily [24]

You must observe the object twice.

-- Look at it the first time, and make a mark where it is.

-- After some time has passed, look at the object again, and
make another mark at the place where it is.

-- At your convenience, take out your ruler, and measure the
distance between the two marks.

What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.

8 0

3 years ago

A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a

3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec

Explanation:

We have given acceleration of the car $a_1=5.9m/sec^2$

Acceleration of the stock car $a_2=3.6m/sec^2$

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So $s_1=s_2$

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So $\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6$

After solving t = 3.56 sec

(b) From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m$

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0

3 years ago

A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

5 0

3 years ago

A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

4 0

3 years ago

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

$W = (44 N)(0.9 m) = 39.6 J$

now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

5 0

2 years ago

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