If you are looking for the answer of h&m where is it from, you’ve got the right page. We have approximately 10 FAQ regarding h&m where is it from. Read it below.

## A robot is program to throw a ball in upward

**Ask: **A robot is program to throw a ball in upward in relation to the equation h = 4 + 12x – x² where h is the height in meters and t is the time in seconds. a. How high, from the ground the ball will be after 4 seconds? b. At how many seconds will the ball reach 40 m above the ground?

**Answer:**

theeeanswerrrr issss letter b

## A car moving with a velocity of 55 km/h accelerates

**Ask: **A car moving with a velocity of 55 km/h accelerates uniformly at 2 m/s squared. Calculate the distance it traveled from the place where it accelerated to the place where the velocity is 72 km/h.

**Answer:**

41.63 meters

**Step-by-step explanation:**

Given:

Initial velocity (Vo) = 55 km/hr

Final velocity (Vf) = 72 km/hr

Acceleration = 2 m/s2

Initial distance (Xo) = 0

[tex]55 frac{km}{hr} times frac{1000 : m}{1 : km} times frac{1 : hr}{60 : mins} times frac{1 : min}{60 : sec} \ = 15.28 frac{m}{s} [/tex]

[tex]72 frac{km}{hr} times frac{1000 : m}{1 : km} times frac{1 : hr}{60 : mins} times frac{1 : min}{60 : sec} \ = 20 frac{m}{s} [/tex]

Find time

Acceleration = Vf – Vo / time

Time = Vf – Vo / Acceleration

Time = (20 – 15.28) m/s / 2 m/s2

Time = 2.36 seconds

Distance formula:

[tex]x = xo + vot + frac{1}{2} a {t}^{2} \ x = 0 +15.28(2.36) + frac{1}{2}(2)( {2.36}^{2}) \ x = 36.06 + 5.57 \ x = 41.63 : meters[/tex]

## Directions: Given the position and velocity, calculate the potential (PE),

**Ask: **Directions: Given the position and velocity, calculate the potential (PE), Kinetic (RE),

and total mechanical energy (E) of the ball then answer the questions that follow

At a maximum height (1.20 m)

h = 1.20 m

PE =

v=0 m/s

KE –

m = 0.620 kg

9 = 9.81 m/s2

1.2m

At 0.60 m

PE

KE

.

h = 0.600 m

V- 3.43 m/s

m = 0.620 kg

g = 9.81 mis

0.6 m

At 0 m

h = 0 m

PE =

KC =

Om

v=4.85 m/s

m = 0.620 kg

g = 9.81 m/s2

Illustration by Daniel C Villanueva

Figure 1. A bail dropped from a piayer

Guide Questions:

1. Analyze the position of the ball as it moves to the ground. Does the total

mechanical energy (E) change? What can you infer from this?

2 You noticed that the poieniiai energy (PE) upon hitting ine ground became zero.

What happened to the energy possessed by the ball?

3. The kinetic energy (KE) of the ball from the maximum height is zero and

changes to 7.30 J just before hitting the ground. Where does this increase in

energy come from?

4. How does the change in height (h) affect the potential energy (PE)?

5. How does the change in velocity (v) affect the kinetic energy (KE)?

**Answer:**

1.

E= 7.60864

PE=7.29864

KE=0.31J

2.

E= 7.296439

PE=3.64932

KE=3.647119

3.

E= 12.983575

KE=5.6916

PE=7.291975

## the height (H) of the ball thrown into the air

**Ask: **the height (H) of the ball thrown into the air with an initial velocity of 9.8m/s from a height of 2 m above the ground is given by the equation H t = -4.9t²+9.8t+2, where t is the time in seconds that the ball has been in the air. HINDI AKO TUMATANGGAP NG KINABUANG NGA TUBAG

**Answer:**

Problem: The height (H) of the ball thrown into the air with an initial velocity of 9.8 m/s from a height of 2 m above the ground is given by the equation H(t)=-4.9t^2+9.8t+2, where t is the time in seconds that the ball has been in the air

Question: Find the value of t to determine the time it takes the ball to reach 4 m

**Step-by-step explanation:**

Problem: The height (H) of the ball thrown into the air with an initial velocity of 9.8 m/s from a height of 2 m above

the ground is given by the equation H(t)=-4.9t^2+9.8t+2, where t is the time in seconds that the ball has been in the air

Question: Find the value of t to determine the time it takes the ball to reach 4 m

- For it, you need to solve this equation
- -4.9t^2+9.8t+2 = 4.

## A car moving with a velocity of 55 km/h accelerates

**Ask: **A car moving with a velocity of 55 km/h accelerates uniformly at 2 m/s2.

a. Calculate the distance it traveled (in meters) from the place where it accelerated to

the place where the velocity is 72 km/h.

b. How long did it take the car to cover such distance?

**Answer:**

Since the distance traveled is S =frac {V _ 1 ^ 2-V _ 0 ^ 2} {2times a}S=

2×a

V

1

2

−V

0

2

, where V_1V

1

is the final speed, V_0V

0

is the initial, aa is the acceleration, then S =frac {72 ^ 2-54 ^ 2} {2times 2} = 567S=

2×2

72

2

−54

2

=567 m;

Since by another formula the distance is S = V _ 0times t +frac {atimes t ^ 2} {2}S=V

0

×t+

2

a×t

2

, we will express from this equation tt :

frac {atimes t^2}{2}+V_0times t – S=0

2

a×t

2

+V

0

×t−S=0 ,

frac {2times t^2}{2}+54times t – 567=0

2

2×t

2

+54×t−567=0 ,

t=9t=9 s.

## a car moving with a velocity of 55 km/h accelerates

**Ask: **a car moving with a velocity of 55 km/h accelerates of 2 m/s squared. Compute the distance traveled from the place where the acceleration had begun to that where the velocity is 72 km/h. What is the time taken to cover the distance?

**Answer:**

only watch this

**Explanation:**

that is albert einstein qiestion so this is it watch that

answer: only 4x⁵ ×c = 1 x ²

## 46 A car moving with a velocity of 55 km/h

**Ask: **46 A car moving with a velocity of 55 km/h accelerates uniformly at 2 m/s. accelerated to the place where the velocity is 72 km/h. (b) How long did it (a) Calculate the distance it traveled (in meters) from the place where it take the car to cover such distance? Show your solutions.

**Explanation **

*KEEP **ON **LEARNING:**D*

## A car moving with a velocity of 55 km/h accelerates

**Ask: **A car moving with a velocity of 55 km/h accelerates uniformly at 2 m/s2.

a. Calculate the distance it traveled (in meters) from the place where it accelerated to

the place where the velocity is 72 km/h.

b. How long did it take the car to cover such distance?

With solution po if you can ty po need kona po ngayon

**Answer:**

hi

**Explanation:**

## 4. Solve. A car moving with a velocity of 55

**Ask: **4. Solve. A car moving with a velocity of 55 km/h accelerates uniformly at the rate of 2 m/s^2. Calculate

A. the distance traveled from the place where the acceleration had begun to that where the velocity is 72 km/h

B. the time taken to cover this distance

**Answer:**

41.63 meters

**Explanation:**

Given:

Initial velocity (Vo) = 55 km/hr

Final velocity (Vf) = 72 km/hr

Acceleration = 2 m/s2

Initial distance (Xo) = 0

Find time

Acceleration = Vf – Vo / time

Time = Vf – Vo / Acceleration

Time = (20 – 15.28) m/s / 2 m/s2

Time = 2.36 seconds

**Answer:**

41.63 meters

Step-by-step explanation:

Given:

Initial velocity (Vo) = 55 km/hr

Final velocity (Vf) = 72 km/hr

Acceleration = 2 m/s2

Initial distance (Xo) = 0

**Explanation:**

### hope it’s help you

## _______________________________________ 2) These areas are filled with seawater during high

**Ask: **_______________________________________

2) These areas are filled with seawater during high tides and are drained during low tides. They are marshy because they are filled with decomposing plant matter.

T

A

L

S

S

E

H

S

R

A

M

_______________________________________

3) These are also called as tidal flats are areas in estuaries where mud from the rivers is deposited.

D

U

M

S

T

A

L

F

_______________________________________ 4) These are areas in estuaries where solid rocks are found.

Y

K

C

O

R

S

E

R

O

H

S

Answer:

1.Salt. Marshes

2.Mud. Stalf

3.Rocky. Shores

Brainliestpls

Not only you can get the answer of **h&m where is it from**, you could also find the answers of the height (H), 4. Solve. A, 46 A car, A car moving, and _______________________________________ 2) These.