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4 years ago

Teams A and B are in a tug-of-war challenge. Team A wins the challenge.

Physics

2 answers:

xeze [42]4 years ago

8 0

Answer:

A the force was in their direction

EastWind [94]4 years ago

4 0

Answer:

A. The net force was in their direction.

Explanation:

In the game of tug of war the two teams will pull the two ends of the rope with different forces.

When the two teams will pull the two ends of the rope then the team who win is the one who pull the rope towards them.

so here in order to win Team A we know that team A has to pull the rope towards them.

So it must have more force towards them.

So net force is always towards team A in order to win the game

So here correct answer will be

A. The net force was in their direction.

You might be interested in

The Moon takes about 27 days to orbit the Earth. Assuming a circular orbit, how fast is it orbiting? Express your answer in km/h

Vinvika [58]

Answer:

3727.24km/h

Explanation:

Hello!

To solve this problem we must first know what is the distance from the earth to the moon, this will be our radius.

=384400Km

Then we find the distance traveled which would be the perimeter of a circle = 2πr, finally to find the speed we divide the distance traveled by 27 days.

Finally we use conversion factor to have the speed in km / h

solving

$perimeter= 2(3.14)r=2(3.14)(384400Km)=2415256.432km\\Speed=\frac{2415256.432km}{27days} *\frac{1day}{24h} =3727.24km/h$

the moon is orbiting at speed of 3727.24km/h

7 0

3 years ago

How many molecules are indicated by writing 2 H2O? A.) 2 B.) 3 C.) 4 D.) 1

Thepotemich [5.8K]

A, 2 molcules
H2O is one molecule, it has 3 atoms (two hydrogen and one oxygen)
2 H2O are two water molecules

7 0

3 years ago

Read 2 more answers

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

3 years ago

A laser emits a beam of light whose photons all have the same frequency. When the beam strikes the surface of a metal, photoelec

seraphim [82]

Answer:

the no. of ejected electrons per second will increase.

Explanation:

In photoelectric effect, when a light is incident on a metal surface it ejects some electrons from the metal surface. The energy of photon of light must be equal to or greater than the work function of that metal. All the extra energy above the work potential appears as the kinetic energy of the ejected electrons. So, greater he energy of photon greater will be the kinetic energy of the ejected electrons.

A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, the no. of ejected electrons per second will increase.

3 0

3 years ago

Otion

Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

N- W = 0

N = W

X axis (radial)

fr = m a

the acceleration in the curve is centripetal

a = $\frac{v^2}{r}$

the friction force has the expression

fr = μ N

we substitute

μ mg = m v²/r

v = $\sqrt{\mu g r}$

we calculate

v = $\sqrt{0.1 \ 9.8 \ 3}$

v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0

3 years ago