# H&m Where Is It From

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## 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?​

## 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. ​

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)?​

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​

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?

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?​

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

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

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

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:

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