All categories

3 years ago

Newton's First Law says that an object at rest will stay at rest, and an object in motion will stay in motion unless:

Physics

1 answer:

Verdich [7]3 years ago

6 0

Answer: Acted on by equal forces in opposite direction

Explanation:

Newton's First Law says that every body continue in its state of rest or constant speed on a straight line unless being acted upon by an external force.

You might be interested in

Which list places the layers of the sun in the correct order from outermost to innermost?

Readme [11.4K]

Answer:

The Sun's layers consist of the following in this order.

1) Corona

2) Transition Region

3) Chromosphere

4) Photosphere

5) Convection Zone

6) Radiative Zone

and last but not least 7) The Core

Hope this helps ;)

6 0

3 years ago

What is the equation for momentum? Please help need answer ASAP.

Leona [35]

Momentum is mass in motion and only applies to objects in motion. It's a term that describes a relationship between the mass and velocity of an object, and we can see this when it is written in equation form, p = mv, where p is momentum, m is mass in kg and v is velocity in m/s.

3 0

3 years ago

Read 2 more answers

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force $F_c$ . In this case, the only force that applies for that is the static frictional force $f_s$ between the tires and the track. Then, we can write that:

$f_s=F_c$

And since $f_s\leq \mu N$ and $F_c=\frac{mv^{2}}{r}$ , we have:

$\mu N\geq \frac{mv^{2}}{r}$

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

$N-mg=0\\\\\\implies N=mg$

Substituting this expression for $N$ and solving for $v$ , we get:

$\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}$

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

$v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s$

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

7 0

3 years ago

Read 2 more answers

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg$

So, the mass that is attached to the spring is 1 kg.

6 0

3 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago