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Mrrafil [7]
3 years ago
5

A man, a distance d=3~\text{m}d=3 m from a target, throws a ball at an angle \theta= 70^\circθ=70 ​∘ ​​ above the horizontal. If

the initial speed of the ball is v=5~\text{m/s}v=5 m/s, what height hh does the ball strike the building?
Physics
1 answer:
lbvjy [14]3 years ago
4 0

Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

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The type of material that will easily accept the flow of electric current is called a(n)
Sunny_sXe [5.5K]

Answer: Conductor

Explanation: Because they do

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3 years ago
You drop a stone down off a bridge. You are able to count to 4.0 seconds when it finally hits the water. How high is the bridge?
mart [117]

Answer:

The height of the bridge is 78.4 m.

Explanation:

Given;

time of the stone motion off the bridge, t = 4.0 s

acceleration due to gravity, g = 9.8 m/s²

The height of the bridge is given by;

h = ut + ¹/₂gt²

where;

u is the initial velocity of the stone, u = 0

h = ¹/₂gt²

h = ¹/₂(9.8)(4)²

h = 78.4 m

Therefore, the height of the bridge is 78.4 m.

7 0
3 years ago
Which organism makes its own food? A mouse B snake C grass D owl
bearhunter [10]
Your answer will be C: grass

NOT A, because a mouse would eat seeds, grass, etc
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6 0
3 years ago
Read 2 more answers
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0
3 years ago
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