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kap26 [50]
3 years ago
5

A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta

l. If the lawn roller develops 69.4 W of power for 33 s, what distance is the roller pushed? Answer in units of m.
Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

          = 107 cos 13.5°= 104 N

Power = \dfrac{work\ done}{time}

69.4 = \dfrac{W}{33}

W = 2290.2 J

Work done= Force x displacement

displacement= \dfrac{2290.2}{104}

                      = 22.02 m

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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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3 years ago
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Nataly_w [17]
The anwser is C flooding
3 0
3 years ago
which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload
Anuta_ua [19.1K]
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5 0
3 years ago
9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
Korolek [52]

<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

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=1.259ms^2

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3 years ago
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