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Allisa [31]
3 years ago
5

How can a high school student have more momentum than a truck

Physics
1 answer:
morpeh [17]3 years ago
6 0

Answer:

When truck is at rest while student is under motion

Explanation:

Since it is obvious that the mass of a truck is more than that of a student, we know that momentum is a product of mass and velocity

P=mv where m represent mass, v is velocity. When the student has more speed than that of truck, he exerts more momentum. The only way a student can exert more momentum is by having more speed while the truck is at rest. In such case, the momentum of truck will be zero while momentum of student will have a value

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A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t
k0ka [10]

Answer:

20 ft

Explanation:

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3 years ago
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What is a
LiRa [457]

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I think there six points

4 0
3 years ago
Betelgeuse is how many times larger than the sun
Aleksandr-060686 [28]
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
3 0
3 years ago
In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having
Orlov [11]

Answer:

R = 5.28  103 km

Explanation:

The definition of density is

              ρ = m / V

              V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

            V = 4/3 π r³

Let's replace

             4/3 π r³ = m / ρ

             R =∛ ¾ m / ρ π

The mass of the planet is

              M = 5.5 Me

              R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

             ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

             ρ = 1.76 10³ kg / m³

Let's calculate

               R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

               R = ∛ 0.14723 10²¹

               R = 0.528 10⁷ m

               R = 0.528 104 km

               R = 5.28  103 km

8 0
3 years ago
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