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marissa [1.9K]
3 years ago
13

What charge are electrons

Physics
2 answers:
TEA [102]3 years ago
6 0

Answer: negative

Explanation: it may seem that neutrons would be negative but neutrons are neutral, protons are positive, and electrons are negative

Colt1911 [192]3 years ago
4 0

Answer:

negative

Explanation:

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PLEASE HELP!!!!!!!! NEED DONE SOON!!!
zhannawk [14.2K]

Answer:

the object is decelerating

4 0
3 years ago
Read 2 more answers
Which two components must a vector quantity
Lemur [1.5K]

Answer:

d. Direction and magnitude

Explanation:

The two components of a vector are its magnitude and direction.

Magnitude is the quantity of the substance

Direction is the path.

  • Other quantities are called scalar quantities.
  • Scalar quantities have only magnitude but no direction.

Examples of vector quantities are velocity, displacement, acceleration.

4 0
3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
1 year ago
A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L
snow_lady [41]

Answer:

Q=81.56\ W/m^2

Explanation:

Given that

T_1= 210 K

T_2= 150 K

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

Q=\sigma (T_1^4-T_2^4)\ W/m^2

So now by putting the values

Q=\sigma (T_1^4-T_2^4)\ W/m^2

Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2

Q=81.56\ W/m^2

So rate of heat transfer per unit area

Q=81.56\ W/m^2

8 0
3 years ago
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