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Helen [10]
3 years ago
13

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

flight C. Increases, goes to zero, then increases D. Depends on who threw it
Physics
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

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A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
Find the weight of a 25 kg table. (Use g= 10 m/s?)
PtichkaEL [24]

Answer:

weight = 25*10 =250 N

Explanation:

g must be given in units of m/s^2

The weight of any type of body will be the product of his mass by the gravity

where:

m =mass [kg]

F = force [N] or [kg*m/s^2]

g = acceleration [m/s^2]

3 0
3 years ago
Read 2 more answers
A book is launched up along the rough incline. Kinetic energy given to a book at initial point is 100 J. Book comes to stop at s
den301095 [7]

Answer:

(A) 60 J

Explanation:

At state 1

KE₁=100 J

At state 2

KE₂ = 0

U₂=80 J

Given that surface is rough so friction force will act in opposite to the direction of motion

Lets take work done by friction = Wfr

From work power energy

Work done by all forces = Change in kinetic energy

Wfr + U₂=ΔKE

Wfr+80 = 100

Wfr= 20 J

Now when book slides from top position then

Wfr+ U = KEf - KEi

-20 + 80 = KEf-0

KEf= 60 J

(A) 60 J

7 0
3 years ago
Atoms containing radioactive nuclei are called
miss Akunina [59]
The correct answer is radioisotopes.
3 0
3 years ago
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