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Gennadij [26K]
3 years ago
14

If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceler

ation? Is the acceleration positive or negative?
Physics
1 answer:
Katen [24]3 years ago
8 0

Answer:

The acceleration would be positive, pointing to the right.

Explanation:

The acceleration tells us how velocity changes every second. If the train stops, this means its velocity must go from a non-zero value to zero.

To reduce velocity, acceleration must point to the opposite direction. Since velocity is negative, acceleration must be positive.

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a deer with a mass of 176 kg is running head-on towards you with a velocity of 19 m/s. you are going north. find the magnitude a
Igoryamba

Momentum = Mass × Velocity

According to this formula,

Momentum of deer = 176 × 19 = 3344 kg•m/s.

Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.

6 0
3 years ago
Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of
Marysya12 [62]
Well they could go down a hill to gain more kinetic energy.
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3 years ago
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3 0
3 years ago
he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
S_A_V [24]

Answer:

207.4 N

Explanation:

The torque \tau  on a body is

\tau = r* F  where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

\tau = rF\sin \theta

Where \theta is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation \tau = rF\sin \theta

\tau = LF\sin \theta

Where L is the length of the wrench.

Making F the subject

F = \frac{\tau }{{L\sin \theta }}

Force required to pull the wrench is given as,

F = \frac{\tau }{{L\sin \theta }}

Substitute 47{\rm{ N}} \cdot {\rm{m}}  for \tau, 25 cm for L, and 115o for \theta  

\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}  

6 0
3 years ago
A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i
guapka [62]

(a)  If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)  if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0
2 years ago
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