Ask question

Ask question

All categories

Veseljchak [2.6K]

2 years ago

6

Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is someti

mes difficult at first for people to understand.
Which statement is the best example of an object and motion that would make it hard for people to accept Newton’s first law?

A rolling ball eventually slows down and comes to a stop.
A wagon must be pushed before it begins to move.
The heavier the load in a cart, the harder the cart is to turn.
A box does not move when pushed equally from opposite sides.

1 answer:

dolphi86 [110]2 years ago

5 0

Answer:

the heavier the load in a cart ,the harder the cart is to turn

You might be interested in

How long will it take a car to go from a complete stop to 44 km/hr if they are accelerating at 5 km/hr?

Aleonysh [2.5K]

<u>Solu</u><u>tion</u><u> </u><u>:</u><u>-</u>

Given that ,

Initial Velocity of car = 44km / hr.
Final Velocity of car = 0km / hr.
Acceleration = -5km/hr².

To Find :

Time taken to stop the car .

So , here use first equⁿ of motion which is ,

$\boxed{\red{\bf\dag v = u + at }}$

where ,

v is final Velocity.
u is Initial velocity.
a is acceleration.
t is time taken.

Now , substituting the respective values ,

$\tt:\implies v=u+at$

$\tt:\implies 0 = 44 + (-5)t$

$\tt:\implies -44 = -5t$

$\tt:\implies t=\dfrac{44}{5}$

$\underline{\boxed{\red{\tt \longmapsto Time\:\:=\:\:8.8hrs.}}}$

$\boxed{\green{\bf\pink{\dag} Hence\:time\:taken\:to\:stop\:the\:car\:is\:8.8hrs.}}$

6 0

3 years ago

What would happen to the planets in the solar system if the sun’s gravitational force were to suddenly disappear?

Mazyrski [523]

I believe the answer would be B.

Hope this helps!

8 0

4 years ago

Read 2 more answers

A spring with a spring constant of k = 185.0 N / m is oriented vertically, with one end on the ground. What distance does the sp

uysha [10]

Answer:

The distance spring compresses (x) = 0.0811 m

Explanation:

Spring constant (k) = 185 N / m

mass (m) = 1.53 kg

When mass is placed upon the spring the spring force is equal to weight of the mass.

⇒ Spring force (F) = weight of object

⇒ Spring force (F) = k × x

And weight of the object = mg

⇒ k x = mg -----------------(1)

Put all the values in equation (1) we get

⇒ 185 × x = 1.53 × 9.81

⇒ x = 0.0811 m

This the distance spring compresses, when mass is placed upon it.

8 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago

Brainliest for best answer

Alina [70]

A. The restoring force is tripled

4 0

3 years ago

Read 2 more answers