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Sergeeva-Olga [200]
3 years ago
10

A sprinter can go from a state of resting to running at 7 m/s in just 3 seconds. What is the sprinter's average acceleration?

Physics
1 answer:
liberstina [14]3 years ago
4 0
The formula for this equation is:
change in velocity/time taken = acceleration

7/3 = 2.3...

so the answer is 2.3 m/s
(I am not positive with the answer but I do know that is the formula for the equation)
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A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch
Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0
3 years ago
Pls help ASAP now! This is physical science
const2013 [10]

Option A is correct answer

6 0
3 years ago
Read 2 more answers
A bicyclist travels $4.5\text{ km}$ west, then travels $6.7\text{ km}$ at an angle $27.0^\circ$ South of West.
MatroZZZ [7]

Answer:

10.88 km

Explanation:

We shall represent displacement in terms of i , j  unit vectors in the direction of east and north .

4.5 km due west

D₁ = - 4.5 i

6.7 km at an angle of 27° south of west

D₂ = - 6.7 cos27 i - 6.7 sin27j

= - 6.7 x .89 i - 6.7 x .45 j

= - 5.96i - 3 j

Total displacement

= D₁ + D₂

=  - 4.5 i - 5.96i - 3 j

= -10.46 i - 3j

Magnitude = √ ( 10.46² + 3²)

= √ ( 109.41 + 9)

= √ 118.41

= 10.88 km .

7 0
2 years ago
At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?
photoshop1234 [79]
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2}  )=12.38
And so
\ln( \frac{K_2}{K_1})=12.38
And using K_1=6.1\cdot 10^{-8} s^{-1} , we find K2:
K_2=K_1 e^{12.38}=0.0145 s^{-1}


5 0
3 years ago
If the magnitude of the electrostatic force felt by a charge is three times that from another charge, but their charges are the
Aleks [24]

Answer:

three times

Explanation:

F=Eq

3F=3Eq

5 0
3 years ago
Read 2 more answers
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