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2 years ago

4

If a ball is given acceleration of 3.0 meters/second2 while being pushed with a force of 0.75 newtons, what is the mass of the b

all?

2 answers:

Gnoma [55]2 years ago

8 0

By definition of Newton's second law we have to:

$F = ma$

Where,

F: Applied force

m: mass of the object

a: acceleration of the object

Clearing the mass we have:

$m = \frac{F}{a}$

Substituting values we have:

$m = \frac{0.75}{3}\\m = 0.25 Kg$

Answer:

The mass of the ball is:

$m = 0.25 Kg$

AlekseyPX2 years ago

5 0

<span>Mass = force/acceleration. Mass = 0.75/3. The mass is therefore 0.25 units.</span>

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Answer:i=300 mA

Explanation:

Given

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$\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}$

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$i_0=\frac{15}{50}=0.3 A$

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$i=0.3\times 0.9998$

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2 years ago

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

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As the energy in the system is conserved we have

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2 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

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Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

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2 years ago

If you want to maximize the magnetic force on a current in a conductor, how should you orient the current relative to the magnet

tia_tia [17]

Answer:

Should place the current perpendicular to the magnetic field

Explanation:

The magnetic force exerted on a current-carrying wire is given by

$F=ILBsin \theta$

where

I is the current in the wire

L is the length of the wire

B is the magnetic field

$\theta$ is the angle between the direction of the wire and the magnetic field

As we see from the formula, the magnetic force is maximum when

$sin \theta=1$

which means

$\theta=90^{\circ}$

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2 years ago

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Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

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distance: $4m$

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horizontal component: $F_{x}=Fcos\theta$

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but from the given information we know that $F_{y}=12N$

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$Fsen\theta =12N$

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the work done is W=173.48J

7 0

2 years ago