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sertanlavr [38]
3 years ago
12

Standing on a bridge, you throw a stone straight upward. The stone hits a stream28.3 m below the point at which you release it2.

70 s later. What is the speed of the stone (in m/s) just before it hits the water? Please do not include any units in your answer below. Type in only the numerical result. If you include units, your answer will be marked as incorrect.
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

MISSING DATA, SPEED AT WHICH IT WAS LAUNCHED OR INITIAL

DISTANCE THAT REACHED UPWARDS

Explanation:

ANYWAY I LEAVE YOU THE LINK, AUÍ AHY MORE INFORMATION ON THE

SUBJECT.

https://gscourses.thinkific.com

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A car moving in the positive direction with an initial velocity of 26 m/s slows down at a constant rate of -3 m/s2. What is its
Greeley [361]
The answer is 5 m/s
- 3 =  \frac{x - 26}{ 7}  \\ x =  \frac{7 \times(  - 3)}{1}  + 26 =  - 21 + 26 \\  = 5




good luck
8 0
3 years ago
What is direct current? In which direction does current go according to the electron flow convention?
KatRina [158]

Answer:

Answer to the question is:

Explanation:

Direct Current:

It is that current where electrons circulate in the same amount and sense in time, that is, flowing in the same direction. Its polarity is invariable and causes a current of relatively constant amplitude to flow through a load. This type of current is known as direct current (DC), and is generated by a battery.

the current of electrons will leave the negative terminal of the battery, (because they repel each other and also repel free electrons in the copper conductor), and go to the positive terminal where there is a lack of electrons, passing through the circuit to which it is connected. In this way the electric current is produced.

4 0
3 years ago
In a “minute to win it” game, cards are placed between cups to stack them. The contestant then pulls the card out in hopes that
Luden [163]

Answer:

There is no friction between the card and the cup.

Explanation:

4 0
3 years ago
Read 2 more answers
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c
marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
  • Distance=s=96m
  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

3 0
3 years ago
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