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Olegator [25]
3 years ago
8

While mowing your lawn, you push a lawnmower 120. m with a constant force of 300. N. How much work have you done, assuming that

all force is in the same direction as the direction of motion? Express your answer in scientific notation.
a)2.5 x 101 J
b)2.50 x 101 J
c)3.6 x 104 J
d)3.60 x 104 J
Physics
1 answer:
Montano1993 [528]3 years ago
3 0
We have: Work done = Force × Displacement
Here, Force = 300 N
Displacement = 120 m

Substitute their values into the formula:
W = 300 × 120 
W = 36000

In scientific notation, it would be: 3.6 × 10⁴

Finally, option C would be your correct answer.

Hope this helps!
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A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. Determine the work done in climbing the hill.​
iren [92.7K]
W=m/d. w=51.7\1.20 W=43.083333333J
6 0
2 years ago
This question is typical on some driver’s license exams: A car moving at 45 km/h skids 15 m with locked brakes. How far will the
sergiy2304 [10]

from kinematics equation if we know that final speed is ZERO and initial speed is given that due to constant deceleration the object will stop in some distance "d" and this distance can be calculated by kinematics

v_f^2 - v_i^2 = 2 a d

0 - v^2 = 2(a)d

here acceleration due to friction will be same at all different speed

so for 45 km/h speed the distance of stop is 15 m

while at other speed 112.5 km/h the distance will be unknown

now we will have

0 - 45^2 = 2(a)15

0 - 112.5^2 = 2(a)d

now divide above two equations

\frac{45^2}{112.5^2} = \frac{15}{d}

d = 93.75 m

So it will stop in distance 93.75 m

8 0
3 years ago
s A horizontal insulating rod of length 11.0-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge
guajiro [1.7K]

Answer:

F = 2.26 ×  10⁻³ N

Explanation:

given,

length of rod = 11 cm

charge  = 19 nC

linear charge density = 3.9 x 10⁻⁷ C/m

electric force at 2 cm away.

E(r) = \dfrac{2K\lambda}{r}

F = E q

F= \dfrac{2K\lambda\ q}{L}\int \dfrac{dr}{r}

integrating from 0.02 to 0.02 + L

F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]

F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]

F = 2.26 ×  10⁻³ N

5 0
4 years ago
Points A and B lie above an infinite plane of negative charge that produces a uniform electric field of strength E=10 N/C. What
olasank [31]

There is no illustration of the problem provided but I'll attempt to provide an answer.

The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:

║E║ = ΔV/d

║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.

We are given the following values:

║E║= 10N/C

d = 3m

Plug these values in and solve for ΔV

10 = ΔV/3

ΔV = 30V

4 0
4 years ago
#2
sladkih [1.3K]

Answer:

mass and distance

Explanation:

force is mass while motion can also be regard as distance or movement

3 0
3 years ago
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