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Xelga [282]
2 years ago
11

11. A car travels at 25 m/s to the North. It has an acceleration of 2 m/s’ to the south

Physics
1 answer:
Anna11 [10]2 years ago
4 0

Answer:

delta x= 100m

Explanation:

vi= 25 meters

a= -27meters

t=20s

find delta x

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What does the m stand for in the enthalpy equation?
Morgarella [4.7K]

Answer:

Use the formula ∆H = m x s x ∆T to solve.

Explanation:

Once you have m, the mass of your reactants, s, the specific heat of your product, and ∆T, the temperature change from your reaction, you are prepared to find the enthalpy of reaction. Simply plug your values into the formula ∆H = m x s x ∆T and multiply to solve.

4 0
2 years ago
Help solve these two problems im having trouble trying to start these problems?​
belka [17]

Answer:

25.  Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km

6 0
3 years ago
The half-life of carbon-14 is 5370 years. The carbon-14 levels in a fossil indicate that 6 half-lives have passed. How old is th
vodomira [7]
On half life is 5370 years; 6 half lives have passed. You just multiply,

5370*6 = 32,220 years
8 0
3 years ago
Read 2 more answers
A block slides on a frictionless, horizontal surface with a speed of 1.32 m/s. The block encounters an unstretched spring and co
Rus_ich [418]

Answer:

Explanation:

The given time is 1 / 4 of the time period

So Time period  of oscillation.

= 4 x .4 =1.6 s

When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum

velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .

So required velocity is the maximum velocity of the block while remaining in contact with the spring.

v ( max ) = w A = 1.32  m /s.

3 0
3 years ago
A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on
makkiz [27]

Answer:

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh =  0.262 N

Now, using the law of conservation of energy

initial energy= final energy

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1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

b) 1/2 mv² + 2fh = 1/2 mv₀²

Vg = 19 m/s

6 0
3 years ago
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