How does a vector quantity differ from a scalar?

A large truck is moving at 22.0 m/s. if its momentum is 125,000 kg • meters per second, what is the truck's mass? 176 kg 2750 kg

solong [7]

The mass of the large truck is determined as 5680 kg.

<h3>Mass of the truck</h3>

The mass of the truck is calculated as follows;

P = mv

where;

P is momentum
m is mass
v is velocity

m = P/v

m = 125000/22

m = 5680 kg

Thus, the mass of the large truck is determined as 5680 kg.

Learn more about momentum here: brainly.com/question/7538238

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7 0

1 year ago

To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m

Lady bird [3.3K]

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

$P_{g}=\rho g h$

Where, $\rho$ =density

h = height

Put the value into the formula

$P_{g}=1040\times9.8\times4.94$

$P_{g}=50348.48\ Pa$

Pressure in atmospheres

$1\ atm =101.3\ kPa$

$P_{g}=\dfrac{50348.48}{101325}$

$P_{g}=0.4969\ atm$

Hence, The minimum gauge pressure is 0.4969 atm.

7 0

3 years ago

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Which list correctly describes the usual order of planets inward toward the sun?

Sati [7]

Option B

Neptune, Uranus, Saturn, Jupiter, Mars, Earth, Venus, Mercury correctly describes the usual order of planets inward toward the sun

<u>Explanation:</u>

Our solar system continues much considerably than the eight planets that revolve around the Sun. The position of the planets in the solar system, commencing inward to the sun is the accompanying: Neptune, Uranus, Saturn, Jupiter, Mars, Earth, Venus, Mercury.

Most next to the Sun, simply rocky material could resist the heat. For this logic, the first four planets: Mercury, Venus, Earth, and Mars are terrestrial planets. The four large outer worlds — Jupiter, Saturn, Uranus, and Neptune: because of their enormous size corresponding to the terrestrial planets. They're also frequently composed of gases like hydrogen, helium, and ammonia preferably than of rocky surfaces.

7 0

2 years ago

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

3 years ago

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ = 400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction, μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

$v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f$

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

$S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2$

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0

3 years ago

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