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Citrus2011 [14]
3 years ago
14

As part of a NASA experiment, golfer Tiger Woods drives a golf ball on the moon, where g = 1.60 m/s2. He launches a golf ball wi

th a speed of 285 km/h, at an angle of 42° with the horizontal. What horizontal distance will his drive travel before landing back on the surface of the moon. Ignore the curvature of the moon.​
Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

Horizontal distance, R = vo2 sin(2x45)/g

v = 285 km/h = 79.17 m/s

R = 79.172 x 1/1.6

R = 3917 m

R = 3.9 km

You might be interested in
An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi
Lynna [10]

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

6 0
3 years ago
1. Simplify the following expression<br> 8-6/4-12+3^2
DanielleElmas [232]

Answer:

-5/2

Explanation:

8-6/4-12+3^2

8-6/4-12+9

2/4-12+9

LCM =4

(2-48+36)/4

(-46+36)/4

-10/4

-5/2

5 0
2 years ago
A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground traveling 5 m/s.
nignag [31]
The kinetic energy (KE) is 250 J and the gravitational potential energy (GPE) is 392 J
4 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = \frac{1}{3} g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

6 0
3 years ago
The rate traveled from Amarillo to Austin by a bus averages 60 miles per hour. The bus arrived in Austin after eight hours of tr
jeyben [28]

Answer:

the car would arrive after 10 hr to Austin.

Explanation:

bus average from amarllo to austin = 60 miles per hour

time takne by bus to reach austin = 8 hr

automobile average from amarllo to austin = 60 miles per hour

from the information given in the question:

60 mph ---    8 h

48 mph ---  x h

By using The inverse variation:

60 : 48 = x : 8

60* 8 = 48* x

480 = 48*x

x = 480/48

x = 10 h

the car would arrive after 10 hr to Austin.

3 0
2 years ago
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