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3 years ago

5

Two soccer players kick the same 1-kg ball at the same time in opposite directions. One kicks with a force of 25 N; the other ki

cks with a force of 10 N. What is the resulting acceleration of the ball, in m/s2 Answer here I

2 answers:

Studentka2010 [4]3 years ago

7 0

Answer:

15 (These are just here to increase the word limit)

andre [41]3 years ago

5 0

Answer:

The forces offset to produce a net force of 8 N

since the ball is 1 kg

F = ma

a = f/m = 8 N/1 kg = 8 (kg m)/(s^2 kg) = 8 m/s^2

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Write two main group of animals

Nesterboy [21]

Answer:

Animals are classifield as two main groups in the animal kingdom : Vertebrates and invertebrates .

Explanation:

keep it up....

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2 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos:

$T = \frac{1}{f}$ (1)

Donde $f$ es la frecuencia, en hertz.

( $f = 4\,hz$ )

$T = \frac{1}{4\,hz}$

$T = 0.25\,s$

Ahora determinamos la fuerza aplicada sobre la prenda ( $F$ ), en newtons:

$F = m\cdot a$ (2)

$F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}$ (2b)

Donde:

$m$ - Masa de la prenda, en kilogramos.

$r$ - Radio interior del tambor, en metros.

( $m = 0.32\,kg$ , $r = 0.4\,m$ , $T = 0.25\,s$ )

$F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}$

$F \approx 80.852\,N$

La velocidad lineal de la lavadora es:

$v = \frac{2\pi\cdot r}{T}$ (3)

( $r = 0.4\,m$ , $T = 0.25\,s$ )

$v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}$

$v \approx 10.053\,\frac{m}{s}$

Y la velocidad angular del tambor de la lavadora:

$\omega = \frac{2\pi}{T}$

( $T = 0.25\,s$ )

$\omega = \frac{2\pi}{0.25\,s}$

$\omega \approx 25.133\,\frac{rad}{s}$

7 0

2 years ago

Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed

Ivanshal [37]

Answer:

Part a)

$f_w = f_g = 4.57 \times 10^{14} Hz$

Part b)

$\lambda_w = 492 nm$

$\lambda_g = 437.3 nm$

Part c)

$v_w = 2.25 \times 10^8 m/s$

$v_g = 2.0 \times 10^8 m/s$

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

$f = \frac{v}{\lambda}$

$f = \frac{3 \times 10^8}{656 \times 10^{-9}}$

$f = 4.57 \times 10^{14} Hz$

Part b)

As we know that the refractive index of water is given as

$\mu_w = 4/3$

so the wavelength in the water medium is given as

$\lambda_w = \frac{\lambda}{\mu_w}$

$\lambda_w = \frac{656 nm}{4/3}$

$\lambda_w = 492 nm$

Similarly the refractive index of glass is given as

$\mu_w = 3/2$

so the wavelength in the glass medium is given as

$\lambda_g = \frac{\lambda}{\mu_g}$

$\lambda_g = \frac{656 nm}{3/2}$

$\lambda_g = 437.3 nm$

Part c)

Speed of the wave in water is given as

$v_w = \frac{c}{\mu_w}$

$v_w = \frac{3 \times 10^8}{4/3}$

$v_w = 2.25 \times 10^8 m/s$

Speed of the wave in glass is given as

$v_g = \frac{c}{\mu_g}$

$v_g = \frac{3 \times 10^8}{3/2}$

$v_g = 2 \times 10^8 m/s$

4 0

3 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

2 years ago

A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se

grigory [225]

First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)

6 0

3 years ago