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3 years ago

Match each letter to the description

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

A woman walks in a straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

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1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of

raketka [301]

Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

6 0

3 years ago

square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch

Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If ' $W$ ' be the work done on an object to change its kinetic energy from an initial value ' $K_{i}$ ' to the final value ' $K_{f}$ ', then mathematically,

$W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)$

where ' $m$ ' is the mass of the object and ' $v_{i}$ ' and ' $v_{f}$ ' be the initial and final velocity of the object respectively. If ' $F_{net}$ ' be the net force applied on the car, as per given problem, and ' $s$ ' is the displacement occurs then we can write,

$W = F_{net}~.~s.......................................................(II)$

Given, $m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m$ .

Equating equations (I) and (II),

$&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N$

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3 years ago

You drop a rock off of a cliff at exactly the edge after 25 seconds exactly you hear a splash your physics friend tells you the

Reptile [31]

The speed of sound is 340.29 meters per second.

Knowing that, we can calculate how high this cliff is by 340.29 * .4

The cliff is 340.29 * .4 = 136.12 meters

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Answer:

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