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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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A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on the top. A large loa
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The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.

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3 years ago
Human eyes detect only a very small band of the electromagnetic spectrum. However, some animals and insects see in higher and lo
hoa [83]

Answer:

Squids = 450 - 490 nm (Moderate Frequency) (Blue)

Bees = 300 - 650 nm (Lower Frequency Bands)

Frogs = 280 - 580 nm (Very Low Frequency)

Explanation:

All of the above mentioned ranges are compared to that of humans.

I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE

3 0
3 years ago
What two factors are a part of thermohaline circulation
Margarita [4]

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

7 0
3 years ago
The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A
steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}

\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}

\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}

\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}

\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}

\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}

Work done is defined as

W = \overrightarrow{F}.\overrightarrow{S}

W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k}  \right )

W = -1120 + 450 + 70

W = - 600 J

3 0
3 years ago
The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

v_i=6.1 m/s

v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}

x=\sqrt{2.7x10^{-3}m^2}=0.052m

5 0
2 years ago
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