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Find the Y component of 35m/s at 57 degrees from the X-axis.

rosijanka [135]

Answer:

35m/s[57o].

X = 35*Cos57 =

Y = 35*sin7 =Explanation:

learn man but there u go

3 0

3 years ago

What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?

vitfil [10]

Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

= 30 N

3 0

2 years ago

What is the answer of A and B question 2

jok3333 [9.3K]

They have the writing

5 0

3 years ago

Read 2 more answers

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Nata [24]

The y-component of the acceleration is $0.28 m/s^2$

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

Here we have:

Initial velocity is $u=2.25 m/s$ at $\theta_1=50.0^{\circ}$ , so its y-component is $u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s$

Final velocity is $v=4.65 m/s$ at $\theta_2=120.0^{\circ}$ , so its y-component is $v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s$

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

$a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

2 years ago