What force is affected when the distance between two objects remains the same and the mass of each object is doubled?.

Blizzard [7]

Answer:

A submission from India says the name Priyadharshini means "Beloved and pleasing to look at" and is of Sanskrit origin. According to a user from India, the name Priyadharshini is of Indian (Sanskrit) origin and means "God gift".

6 0

3 years ago

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1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es

bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

$v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h$

or

$v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s$

So, the required speed is 15 km/h or 4.16 m/s.

7 0

3 years ago

Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th

vitfil [10]

Answer:

$q=7.65*10^{-8}C$

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

$F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N$

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

$F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C$

8 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not

stiks02 [169]

Answer:

2

Explanation:

2

5 0

3 years ago

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