All categories

3 years ago

Which of the following equations describe the 2nd law of motion?

Physics

2 answers:

Paladinen [302]3 years ago

7 0

Answer:

b. force= mass *acceleration

Explanation:

Sever21 [200]3 years ago

4 0

Answer:

b. F = m × a is the correct answer.

You might be interested in

A 1000 kg car accelerates from rest at a rate of 10 m/s² for 3 seconds. A) what is the final velocity of the car?

Lorico [155]

Answer:

Refer to the attachment!~

5 0

3 years ago

Milk is an example of a .

Verizon [17]

Answer: Homogenous mixture.

Explanation:

3 0

3 years ago

Read 2 more answers

The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i

shtirl [24]

Answer: The coefficient of kinetic friction is μ = 0.6

Explanation:

For an object of mass M, the weight is:

W = M*g

where g is the gravitational acceleration: g = 9.8m/s^2

And the friction force between this object and the surface can be written as:

F = W*μ

where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)

In this case, the weight is:

W = 20N

And the friction force is:

F = 12N

Replacing these values in the equation for the friction force we get:

12N = 20N*μ

(12N/20N) = μ = 0.6

The coefficient of kinetic friction is μ = 0.6

7 0

3 years ago

Read 2 more answers

A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque

Marianna [84]

Answer:

132

Explanation:

if you round it is correct

7 0

3 years ago

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago