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ArbitrLikvidat [17]
3 years ago
6

1. Define friction. Name and describe two kinds of friction.

Physics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

Friction is a force that holds back the movement of a sliding object.

Explanation:

The two types of friction: Static friction and Kinetic friction.  Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.

You might be interested in
Why is figure 5 an unhelpful visualization tool for this data set? <br><br> Please help!
Paraphin [41]

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

8 0
2 years ago
Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?
OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0
2 years ago
Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of
mario62 [17]
I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³. 
I'm not sure whether that is the answer you were looking for, but that's what I found.
5 0
2 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
3 years ago
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