An object accelerates at 20 m/s2. By how much does the speed change in one second?

Which of the following renaissance scientists made improvements to the telescope?

LenaWriter [7]

Michelaneglo DDDDDDDDDDDDDDDDDDDDDDDDD

7 0

3 years ago

As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds

lora16 [44]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-a*t$

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

$10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]$

Note: The negative sign in the above equation means that the velecity is decreasing.

2)

To solve this second part we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2} -2*a*x\\$

where:

x = distance [m]

$(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]$

7 0

2 years ago

Initially (at time t = 0) a particle is moving vertically at 7.5 m/s and horizontally at 0 m/s. Its horizontal acceleration is 1

Lelu [443]

Answer:

t = 0.657 s

Explanation:

given,

initial vertical velocity = 7.5 m/s

initial horizontal velocity = 0 m/s

angle = 49◦

using kinetic equation

final velocity in vertical direction

v sinθ = u_y - gt ........................(1)

final velocity in horizontal direction

v cosθ = u_x + a_x × t

here u_x = 0.0 m/s

v cosθ = a_x×t ......................(2)

Dividing equation (1) / (2)

$tan \theta =\dfrac{u_y - gt}{a_x\times t}$

solving for time t

$t = \dfrac{u_y}{tan \theta \times a_x + g}$

u_y = initial velocity along x direction

acceleration along a_x = 1.4 m/s²

g = acceleration due to gravity = 9.8 m/s²

θ = 43° , u_y = 7.5 m/s

$t = \dfrac{7.5}{tan 49^0\times 1.4+ 9.8}$

t = 0.657 s

time taken by the particle is t = 0.657 s

6 0

2 years ago

Consider the two vectors A = 3 î − ĵ and B = − î − 2 ĵ.

LUCKY_DIMON [66]

Answer:

(a) A+B = 2i-3j

(B) A-B = 4i + j

Explanation:

We have given two vectors A = 3i-j and B = -1-2j

We have to find the two vectors that is A+B and A-B

(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B

So A+B = 3i-j-i-2j = 2i-3j

(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j

6 0

3 years ago

A cat is on a merry-go-round that completes 1 full rotation in 6 seconds. The cat sits at a radius of 8.4 metres from the centre

madam [21]

To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.

The velocity from the uniform circular motion can be described as

$v = \frac{2 \pi r}{T}$