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2 years ago

4

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

1 answer:

Mila [183]2 years ago

3 0

Answer:

C) from 4.5 to 5.0 seconds

Explanation:

The graph shows the distance of the ball from the initial point (y-axis) against the time (x-axis).

In such graph, the speed of the ball corresponds to the slope of the curve, since it is equal to the ratio between the distance travelled (increment in the y-axis) and time taken (increment in the x-axis):

$v=\frac{\Delta y}{\Delta x}=\frac{d}{t}$

Therefore, we can say that the object is moving when the slope of the curve is not zero: when the slope of the curve is zero, instead, the object is at rest.

From the graph, we observe that between t=4.5 s and t=5.0 s the slope of the curve is zero, because the curve is horizontal. This means that the speed is zero, so the object is not moving.

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Bears eat fruits such as berries and animals such as fish. They hibernate in the winter. They give birth to live young . Which o

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Answer:

Bears are consumers

Explanation:

7 0

2 years ago

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago

A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals

spin [16.1K]

Answer:

$\frac{v_{cm}}{\omega} = 1.122\cdot R$

Explanation:

According to the statement of the problems, the following identity exists:

$K_{t} = K_{r}$

$\frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}$

After some algebraic handling, the ratio is obtained:

$\frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}$

$\frac{v_{cm}}{\omega} = 1.122\cdot R$

4 0

2 years ago

An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac

Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

= (50 - 0)/4

= 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0

2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

2 years ago