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yKpoI14uk [10]
3 years ago
15

When a cup is placed on a table, which force prevents the cup from falling to the ground?

Physics
2 answers:
pshichka [43]3 years ago
5 0
I believe it is normal force.
disa [49]3 years ago
3 0

Answer: Normal force prevents the cup from falling to the ground.

Explanation :

When a cup is placed on a table, the forces acting on the cup are the gravitational force and the normal force. The gravitational force acts in the downward direction while the normal force acts in an upward direction.

In the figure, N is the normal force and mg is the weight of an object acting in downward direction.

It is clear that the gravitational force is balanced by the normal force.

So, the correct option is (b) " normal force ".

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The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr
SVETLANKA909090 [29]

Answer:

∅=63.43^{0}

Explanation:

z=impedance

x_{l}=2R

R=R

The resultant of the resistances in the circuit is called impedance

x_{l} is inductive reactance of the circuit

R is the resistance of the resistor

z=\sqrt{xl^{2}+R^2 }

z=\sqrt{2R^{2}+R^2 }

Z=\sqrt{5R^2}

Z=R\sqrt{5}ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅=63.43^{0}

phase angle is ∅=63.43^{0}

3 0
3 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

8 0
3 years ago
Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max
yanalaym [24]

Answer:

\theta=39.49^{\circ}

Explanation:

Maximum height of the pumpkin, H_{max}=9.99\ m

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

H_{max}=\dfrac{v^2\ sin^2\theta}{2g}

On rearranging the above equation, to find the angle as :

\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})

\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})

\theta=39.49^{\circ}

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0
3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

6 0
3 years ago
a thermometer has its steam marked in millimetre instead of degree celsius. The lower fixed points is 180 mm calculate the tempe
Nady [450]

Answer:hi

Explanation:

4 0
3 years ago
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