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Ierofanga [76]
7 months ago
10

a projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. what

is the projectile's horizontal speed after it has fallen 25 meters? 22 m/s 35 m/s
Physics
2 answers:
STatiana [176]7 months ago
5 0

Answer: 35m/s

Explanation: answer from castle learning

QveST [7]7 months ago
3 0

A projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. projectile's horizontal speed after it has fallen 25 meters is 35 m/s.

<h3>what is speed ?</h3>

Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

The formula of speed is represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

There are four types of speed such as Uniform speed, Variable speed, Average speed, Instantaneous speed

Uniform speed is an uniform speed when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the uniform speed the ratio of total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as an object is moving with variable speed, then the speed at any instant of time is known as instantaneous speed.

For more details regarding speed, visit

brainly.com/question/13263542

#SPJ2

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One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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