All categories

3 years ago

What is the SI unit for speed and velocity?

Physics

2 answers:

rosijanka [135]3 years ago

5 0

Answer:

Meter per second

Explanation:

Velocity and speed use the same units when being measured. The SI unit of distance and displacement is the meter. The SI unit of time is the second. The SI unit of speed and velocity is the ratio — the meter per second .

Anarel [89]3 years ago

4 0

Answer:

the answer is meter per second

You might be interested in

How much work is done on a 75 newton bowling ball when you carry it horizontally across a 10 meter room

svlad2 [7]

F = force applied to hold the weight of the bowling ball = weight of the bowling ball = 75 N

d = distance through which the bowling ball is moved horizontally = 10 meter

θ = angle between the force in vertically upward direction and displacement in horizontal direction = 90

W = work done on the bowling ball

work done on the bowling ball is given as

W = F d Cosθ

inserting the values

W = (75) (10) Cos90

W = (75) (10) (0)

W = 0 J

6 0

2 years ago

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago

3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

5 0

2 years ago

Read 2 more answers

What accounts for most of the mass in the universe?

aev [14]

Around 80 percent of the mass of the universe is made up material known as "Dark matter". It does not emit light or energy but the influence of it can be detected or observed gravitationally. Motions of stars and galaxy tell us how much mater there is, but somehow the speed of rotation of galaxy does not add up to its mass alone, there is a certain amount of matter really not accounted for. Dark matter maybe made up of non-baryonic matter, or perhaps what scientist called the WIMPS or (weakly interacting massive particles.)

5 0

3 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago