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adell [148]
2 years ago
15

A 10 kg wagon is accelerated by a constant force of 60 N from an initial velocity of 5.0 m/s to a final velocity of 11 m/s. What

is the impulse received by the wagon? a)15 N s b)60 N s c)17 N s d)70 N s
Physics
1 answer:
Readme [11.4K]2 years ago
3 0
The answer is 274774
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A concave lens can only form a. A. real image. B. reversed image. C. virtual image. D. magnified image.
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A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
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3 years ago
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Which kingdom includes only organisms that are heterotophic
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Kingdom Animalia includes only organisms that are heterotrophic.
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2 years ago
Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C
Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo  Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

6 0
2 years ago
Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion s=ut+\frac{1}{2}at^2

40=0\times 10+\frac{1}{2}a10^2

a=0.8944m/sec^2

Now from first equation of motion v=u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So v=u+at=0+0.8944\times 10=8.944m/sec

6 0
3 years ago
Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and coll
Len [333]

Answer:

Explanation:

initial velocity v = 2.1  x 10⁷ m/s

acceleration a = 5.1 x 10¹⁵ m /s²

horizontal distance covered = 5.5 x 10⁻² m

time taken to cover horizontal distance =  5.5 x 10⁻² / 2.1  x 10⁷

= 2.62 x 10⁻⁹ s .

b )

vertical distance travelled due to vertical acceleration

= 1/2 a t²

= .5 x 5.1 x 10¹⁵ x (2.62 x 10⁻⁹)²

= 17.5 x 10⁻³ m

4 0
3 years ago
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