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Len [333]
3 years ago
5

A cyclist maintains a constant velocity of 5 m/s headed away from point A. At some initial time, the cyclist is 247 m from point

A. What will his displacement be in meters from his starting position after 52 seconds?
​
Physics
1 answer:
worty [1.4K]3 years ago
5 0

260 meters im pretty sure because rate*time=distance

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A photon has 3.4 × 10–18 joules of energy. Planck’s constant is 6.63 × 10–34 J•s.
hodyreva [135]
Ok i will answer for real this time. Please give me brainliest.
<span>The Answerr is:
5.12*10^15. Since e=h*f, f=e/h. 3.4*10^(-18)/h.
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5 0
3 years ago
Take the mechanical equivalent of heat as 4 J/cal. A 10-g bullet moving at 2000 m/s plunges
klasskru [66]

Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the material.

The energy of the bullet E = 1/2 mv²

E = 1/2 x 10 x 10⁻³ x (2000)²

E = 2 x 10⁴ J

This heat is used in heating the paraffin

E = m x c ΔT = m x c (Tfinal -Tinitial)

2 x 10⁴ J = 1 x 2.8 x 10³ x (Tfinal -20)

Tfinal = 27.1°C

Thus, the final temperature is 27.1 degree C. The correct option is D.

Learn more about temperature.

brainly.com/question/15267055

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4 0
2 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
Which element has the greatest number of valence electrons?
In-s [12.5K]

Answer:

B. Chlorine

Explanation:

8 0
3 years ago
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Please answer the questions here in the attachment I attached below. They are all multiple choice.
Radda [10]

Answer:

abcbc

Explanation:

8 0
4 years ago
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