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KengaRu [80]
3 years ago
11

Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or

der to get the student-student-skateboard system to accelerate, Student X claims that Student Y should apply a force on Student X while both students stand on the skateboard. Which of the following statements is true regarding the claim made by Student X? a)The claim is correct because Newton’s second law states that an object will accelerate if a net force is applied to the object. b)The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system. c)The claim is incorrect because Student Y cannot apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system. d)The claim is incorrect because both students are internal to the student-student-skateboard system, and internal forces within a system cannot cause the system to accelerate.
Physics
2 answers:
OleMash [197]3 years ago
6 0

Answer:

the answer is B.

Explanation:

The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system

stellarik [79]3 years ago
5 0

Answer:

The claim is incorrect because both students are internal to the student-student-skateboard system, and internal forces within a system cannot cause the system to accelerate.

Explanation:

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A bus traveled down Interstate 35 at a speed of 75 km/h for 4 hours. How many kilometers was the bus speeding?
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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
Masja [62]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

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3 years ago
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