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3 years ago

Which requires more work, carrying a 420 N knapsack up a 200 m hill or carrying a 210 knapsack up a 400 m hill? Why?

Physics

1 answer:

Katen [24]3 years ago

4 0

Answer:

Explanation:

because the weight/work increases bcz of the time period

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Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil

77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is = -2.8 m/s²

6 0

3 years ago

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

Dominik [7]

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

=> $y_c = 1.063 \ m$

3 0

3 years ago

Jeffery gains super strength and pushes two different objects with the same amount of force. Object A accelerates at 40 m/s2, an

Montano1993 [528]

Answer: Object B

Explanation: Acceleration is directly proportional to force and inversely proportional to mass. It implies that more massive objects accelerates at a slower rate.

5 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 3.00

alexira [117]

Energy of a wave:
E = nhc/λ
3000 = (n x 6.63 x 10⁻³⁴ x 3 x 10⁸)/(510 x 10⁻⁹)
n = 7.69 x 10 ²¹ photons per second per meter²
2.70 cm² = 2.70/10,000 m²
= 2.7 x 10⁻⁴
Photons per second = 7.69 x 10 ²¹ x 2.7 x 10⁻⁴
= 2.08 x 10¹⁸ photons per second

4 0

3 years ago

When drawing ray diagrams involving thin lenses, how many rays (at a minimum) are needed show the image distance and magnificati

Rufina [12.5K]

Answer:

GGG he fggfggfufbvg I Rd cbh

6 0

3 years ago