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Basile [38]
3 years ago
5

Serving at a speed of 164 km/h, a tennis player hits the ball at a height of 2.23 m and an angle θ below the horizontal. The ser

vice line is 11.6 m from the net, which is 0.99 m high. What is the angle θ in degrees such that the ball just crosses the net? Give a positive value for the angle.

Physics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

The angle θ is 6.1° below the horizontal.

Explanation:

Please, see the figure for a description of the situation.

The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).

We know the magnitude of these vectors:

magnitude rx = 11.6 m

magnitude ry = 2.23 m - 0.99 m = 1.24 m

Then:

rx = (11. 6 m, 0)

ry = (0, -1.24 m)

r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)

Using trigonometry of right triangles:

magnitude rx = r * cos θ = 11. 6 m

magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m

where r is the magnitude of the vector r

magnitude of vector r:

r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m

Then:

cos θ = 11.6 m / 11.667 m

θ = 6.1°

Using ry, we should obtain the same value of θ:

sin θ = 1.24 m/ 11.667 m

θ = 6.1°

( the exact value is obtained if we do not round the module of r)

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3. An electric motor is used to lift a 6.0 kg of mass through a height of 1 metre. The energy it uses is measure on an energy me
olga55 [171]

Answer:

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.



Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 Jmgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 J

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the

5 0
3 years ago
Explain why semiconductors, or metalloids, deserve their name.
Marysya12 [62]

Answer:

Because their properties like conductivity, electronic configuration and ionization lies in between the metals and nonmetals.

Explanation:

There are a total of six elements that fall in the category of semiconductors.

Namely these are boron, silicon, germanium, arsenic, antimony, and tellurium.

These elements look like metals i.e. are lustrous but do not conduct electricity  so well like a metal does.

Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form mono-atomic anions.

8 0
4 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
4 years ago
What is the speed of a transverse wave in a rope of length 2.00 m and mass 60.0 g under a tension of 500 n?
drek231 [11]

The formula we can use in this case would be:

v = sqrt (T / (m / l))

Where,

v = is the velocity of the transverse wave = unknown (?)

T = is the tension on the rope = 500 N

m = is the mass of the rope = 60.0 g = 0.06 kg

 l = is the length of the rope = 2.00 m

Substituting the given values into the equation to search for the speed v:
v = sqrt (500 N/(0.06 kg /2 m)) 
v = sqrt (500 * 2 / 0.06) 
v = sqrt (16,666.67) 
<span>v = 129.10 m/s</span>

7 0
3 years ago
Junior slides across home plate during a baseball game. If he has a mass of 115 kg, and the coefficient of kinetic friction betw
tatyana61 [14]

Weight Force of Junior = m g = 115kg x 9.81 m/s^2 = 1128.15N then compute for the friction force


Friction Force= WF x (coefficient of kinetic friction) = 1128.15N x 0.35 =  394.8525N or 395N

 

But you can compute in a straightway:

Solution:

= 115 x 9.81 x 0.35

= 394.85

= 395 N

 

It will still give the same results.

4 0
3 years ago
Read 2 more answers
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