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3 years ago

The tenancy of a moving object to continue moving in a strait line or a stationary object to remain in place is called

1 answer:

8 0

The tendency of a moving object to continue moving in a straight line or a stationary object to remain in place is called inertia.

According to law of inertia, an object maintains its state of rest or motion unless acted upon by an external force. an object can not move on its own unless force is applied on it to move it. also an object can not come to stop on its own unless acted upon by an external force to stop it.This is also known as newton's first law.

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The angular velocity of the disk is defined by ω = ( 5 t 2 + 2 ) r a d / s , where t is in seconds. Determine the magnitudes of

ohaa [14]

Answer

given,

Assume radius of the disk be = 0.8 m

angular velocity of disk

ω = ( 5 t² + 2 ) r a d / s

magnitude of velocity and acceleration = ?

At the instant of time, t = 0.5 s

ω = ( 5 x (0.5)² + 2 ) r a d / s

ω = 3.25 r a d / s

$\alpha = \dfrac{d\omega}{dt}$

$\alpha = \dfrac{d}{dt}( 5 t² + 2)$

$\alpha =10 t$

$\alpha =10\times 0.5$

α = 5 rad/s²

velocity

v = ω r

v = 3.25 x 0.8

v = 2.6 m/s

tangential acceleration

$a_t = \alpha r$

$a_t =5 \times 0.8$

$a_t =4\ m/s^2$

normal acceleration

$a_n = \omega^2\ r$

$a_n = 3.25^2\times 0.8$

$a_n =8.45 \m/s^2$

$a = \sqrt{a_n^2 + a_t^2}$

magnitude of the acceleration

$a = \sqrt{8.45^2 + 4^2}$

a = 9.35 m/s²

6 0

3 years ago

What happens to solar radiation when it is absorbed

Zanzabum

Answer:

Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere

8 0

3 years ago

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago

peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie pla

zavuch27 [327]

Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let $\theta$ is the angular distance that the pie plate has moved through.

It is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{183}{4.5}$

$\theta=40.66\ radian$

Since, 1 radian = 57.29 degrees

$\theta=2329.64\ degrees$

Since, 1 radian = 0.159155 revolution

$\theta=6.47124\ revolution$

Hence, this is the required solution.

5 0

3 years ago

A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st

coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0

3 years ago