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Naddik [55]
3 years ago
7

A golfer on a level fairway hits a ball at an angle of 21° to the horizontal that travels 99 yd before striking the ground. He t

hen hits another ball from the same spot with the same speed, but at a different angle. This ball also travels 99 yd. At what angle was the second ball hit? (Neglect air resistance.)
Physics
1 answer:
Delvig [45]3 years ago
3 0

Answer:

69°

Explanation:

Projectiles that land at the same elevation they're launched from will have the same range if the launch angles are complementary (add up to 90°).

The complement of 21° is 69°.

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3. When a person is outside of the system and they add energy to the
NISA [10]

Answer:

We show added energy to a system as +Q or -W

Explanation:

The first law of thermodynamics states that, in an isolated system, energy can neither be created nor be destroyed;

Energy is added to the internal energy of a system as either work energy or heat energy as follows;

ΔU = Q - W

Therefore, when energy is added as heat energy to a system, we show the energy as positive Q (+Q), when energy is added to the system in the form of work, we show the energy as minus W (-W).

5 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
The speed of light in air is 3 x 108 m/s. The speed of light in ice is 2.29 x 108 m/s. What is the refractive index from air to
Studentka2010 [4]

Answer:

η = 1.31

Explanation:

The formula for the refractive index of from air to some other medium is given by the following formula:

\eta = \frac{c}{v}\\

where,

η = refractive index = ?

c = speed of light in air = 3 x 10⁸ m/s

v = speed of light in ice = 2.29 x 10⁸ m/s

Therefore, using these values in the equation we get:

\eta = \frac{3\ x\ 10^8\ m/s}{2.29\ x\ 10^8\ m/s} \\

<u>η = 1.31</u>

4 0
2 years ago
PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY
m_a_m_a [10]

Hello,

<u>Solution for A:</u>

Force = 3.00N

Mass = 0.50 Kgs

Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2

<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

Mass = 0.50Kg

Time = 3.00 Seconds

Again; F = MA (Where F is force, M is mass and A is acceleration)

1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

Velocity = Acceleration * Time = 2 * 3 = 6 m/s

3 0
3 years ago
Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo
icang [17]

Answer:

a

Explanation:

4 0
3 years ago
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