Answer: The force was 13.92 Newtons.
Explanation:
First, let's recall the second Newton's law:
The net force is equal to the mass times the acceleration, or:
F = m*a
where:
F = force
m = mass
a = acceleration.
When the player hits the ball with the bat, he applies a force that accelerates the ball for a small period of time, that increases greatly the speed of the ball.
In this case, we know that:
the mass of the ball is 0.145 kg
The acceleration of the ball is 96m/s^2
Then we can input those values in the above equation to find the force.
F = 0.145kg*96m/s^2 = 13.92 N
The force was 13.92 Newtons.
Superman strikes a golf ball on the ground at a 38 degree angle above the horizontal at 147 m/s. What is the maximum height the golf ball reaches?
255 m
418 m
687 m
1103 m
The ball is hit with an initial vertical velocity of
(147 m/s) sin(38º) ≈ 90.5 m/s
Recall that
v ² - u ² = 2 a ∆y
where u and v are initial and final velocities, respectively; a is acceleration; and ∆y is displacement.
Vertically, the ball is in freefall, so it is only subject to acceleration due to gravity, with magnitude g = 9.80 m/s² in the downward direction. At its maximum height, the ball has zero vertical velocity (v = 0) and the displacement is equal to the maximum height, so
0² - (90.5 m/s)² = 2 (-g) ∆y
∆y = (90.5 m/s)² / (2g)
∆y ≈ 418 m
can someone help im not sure of my answer
Answer:
Yes
Explanation:
But I feel also 2 is correct but your answer is right
A ball rotates on a string as shown below.
If the string in the image above breaks at point 1, which of the following images best shows the direction the ball will travel?
Question 8 options:
Diagram A
Diagram B
Diagram C
Diagram D
Diagram E
Answer:
Diagram A
Explanation:
When an object is travelling in a circular path it has an acceleration called centripetal acceleration, even if it is moving with uniform speed. This acceleration is attributed to the object due to change in the direction of velocity at every instant during the motion.
The direction of the velocity of the object is tangent to the circular path at every instant and if the object beaks free of the circular motion, such as breaking of the string, then it will go in the direction tangent to the circle at that instant. Hence, the diagram that best shows the direction ball will travel is:
Diagram A
A barge is pulled by two tugboats. If the resultant of the forces exerted by the
tugboats is 5kN force directed along the axis of the barge, determine the tension in
each of the ropes for α = 45
Answer:
Approximately [tex]3.5\; \rm kN[/tex] in each of the two ropes.
Explanation:
Let [tex]F_1[/tex] and [tex]F_2[/tex] denote the tension in each of the two ropes.
Refer to the diagram attached. The tension force in each rope may be decomposed in two directions that are normal to one another.
The first direction is parallel to resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \cos(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \cos(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: [tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}}[/tex]. That force should be equal to [tex]5\; \rm kN[/tex].
The second direction is perpendicular to the resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \sin(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \sin(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: [tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}}[/tex]. However, the net force on the barge is normal to this direction. Therefore, the resultant force should be equal to zero.
That gives a system of two equations and two unknowns:
[tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}} = 5\; \rm kN[/tex], and[tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}} = 0[/tex].The second equation suggests that [tex]F_1 = F_2[/tex]. Hence, replace the [tex]F_2[/tex] in the first equation with [tex]F_1[/tex] and solve for [tex]F_1\![/tex]:
[tex]F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN[/tex].
Because [tex]F_1 = F_2[/tex] (as seen in the second equation,) [tex]F_2 = F_1 \approx 3.5\; \rm kN[/tex].
In other words, the tension in each of the two ropes is approximately [tex]3.5\; \rm kN[/tex].
The tension in each of the rope is 3,535.5 N
The given expression:
the resultant force, R = 5 kN = 5000 N
the angle in between the forces, α = 45
To find:
the tension in each of the ropesThe tension in each of the ropes is calculated as follows;
The tension in vertical direction
[tex]F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N[/tex]
The tension in horizontal direction;
[tex]F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N[/tex]
Thus, the tension in each of the rope is 3,535.5 N
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A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1, find the vertical and horizontal component of velocity.
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
[tex]v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h][/tex]
In order to find the vertical component, we must use the sine function of the angle.
[tex]V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h][/tex]
А A pool of water of refractive index
4/3! is 60cm deep. Find its apparent
depth when viewed vertically through
air.
Answer:
Apparent depth = 45 cm
Explanation:
The refractive index of water in a pool, n = 4/3
Real depth, d = 60 cm
We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm[/tex]
So, the apparent depth is 45 cm.
Lauren walks 100m in half a minute. What must her speed have been to travel this distance?
Answer:
3.33 m/s
Explanation:
Using the formula: s = d / t (where s is speed, d is distance, and t is time)
[Convert minutes to seconds then solve]
Half a minute is 30 seconds. Therefore:
s = 100 / 30 = 3.33 m/s
how much heat is necessary to warm 500g of water from 20°c to 65°c
Explanation:
so sorry
don't know but please mark me as brainliest please
feather is dropped from 13 m high after it has fallen 3 m what is its acceleration
Answer:
Explanation:
The constant acceleration of any object, neglecting air resistance and friction, is always -9.8. Proof:
[tex]s(t)=-4.9t^2+13[/tex] which is the position of this feather before it is dropped. The first derivative of position is velocity, so the velocity function for the feather is:
[tex]v(t)=-9.8t[/tex]. We could find the velocity that the feather is experiencing at 3 seconds, but that is not what we are being asked. What we are being asked is the acceleration of the feather at 3 seconds. So we find the acceleration function of the feather:
a(t) = -9.8
It turns out that the time doesn't matter to acceleration due to gravity because feathers and elephants all fall under the same pull.
An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC
where AB is smooth but BC is rough.
a) Compute the mass of the object if the object moves with an acceleration of 5
ms-1 along AB.
Answer:
m = 15.15 kg
Explanation:
Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:
Fx = ma
F Cosθ = ma
where,
F = Magnitude of Force = 85 N
θ = Angle with horizontal = 27°
m = mass of object = ?
a = acceleration of object = 5 m/s²
Therefore,
85 N Cos 27° = m(5 m/s²)
m = 75.73 N/5 m/s²
m = 15.15 kg