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Step2247 [10]
3 years ago
5

What other forces affect the soccer ball after the student's foot stops touching it?

Physics
2 answers:
hodyreva [135]3 years ago
8 0
Gravity, acceleration, velocity
svet-max [94.6K]3 years ago
4 0
Gravity I hope this helps you I'm not very good at this kind of stuff
You might be interested in
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
inysia [295]
The answer is B. 2.45 N/m.
6 0
3 years ago
¿Qué significa que un automóvil viaje con una rapidez de 45[(Km.)⁄(Hr.)]?
netineya [11]

Responder:

Compruebe amablemente la explicación

Explicación:

¿Qué significa que un automóvil viaje con una rapidez de 45 [(Km.) ⁄ (Hr.)]?

Si se dice que un automóvil viaja a una velocidad de 45 km / h, entonces el automóvil cubre una distancia de 45 km en una hora. 45 km / h es una derivación de la velocidad de desplazamiento obtenida de la relación;

Velocidad = distancia / tiempo

Velocidad = 45 km / 1 h.

Distancia = velocidad × tiempo

Distancia = 45 km / h × 1 h

Distancia = 45 km

Por lo tanto, a esa velocidad, se cubrirá una distancia de 45 km por cada hora que se mueva.

4 0
3 years ago
determine the weight in newtons of a woman whose weight in pounds is 125. also, find her mass in slugs and in kilograms. DEtermi
xxMikexx [17]

Answer:

124 just subtract only one

Explanation:

4 0
3 years ago
A centrifugal pump rotates at n˙ = 740 rpm. Suppose the pump has some swirl at the inlet (α1 = 10°) and exits at an angle of 35°
Blizzard [7]

Answer:

Net head = 380cm

bhp = 17.710kW

Explanation:

Angular velocity of centrifugal pump:

w=\frac{2\pi n}{60}=\frac{2\pi (750)}{60}=78.54\frac{rad}{s}

Normal velocity component at outlet of pump:

V_{2,n}=\frac{V}{2\pi r_{2}b_{2}}=\frac{0.573}{2\pi (0.24)(0.162)}}=2.346\frac{m}{s}

Tangential velocity component at exit of the pump:

V_{2,t}=v_{2,n}tan\alpha _{2}=(2.346)tan(35)=1.643\frac{m}{s}

Normal velocity component at inlet of pump:

V_{1,n}=\frac{V}{2\pi r_{1} b_{1}}=\frac{0.573}{2\pi (0.12)(0.18)}=4.22\frac{m}{s}

Tangential velocity component at inlet of the pump:

V_{1,t}=v_{1,n}tan\alpha _{1}=(4.22)tan(0)=0\frac{m}{s}

Equivalent head in centimetre of water column:

H_{water}=H(\frac{rho_{air}}{rho_{water}})\\\\H_{water}=(\frac{w}{g} )(r_{2}V_{2,t}-r_{1}V_{1,t})(\frac{rho_{air}}{rho_{water}}) \\\\H_{water}=(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0)})(\frac{1.2}{998})=38m=380cm

Break horse power:

bhp=rho_{water} gHV=rho_{water} g[(\frac{w}{g})(r_{2}V_{2,t}-r_{1}V_{1,t})]V\\\\bhp=(998)(9.81)[(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0))](0.573)=17710W=17.710kW

8 0
3 years ago
What is the volume of an object with a mass of 15 grams and a density<br> of 2 g/ml
OleMash [197]

Answer: 7.5 ml

Explanation: Density= mass/ volume.

So, volume = mass / density

                    = 15g / 2 g/ml

                    = 7.5 ml

7 0
3 years ago
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