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3 years ago

A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?

Physics

1 answer:

choli [55]3 years ago

3 0

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

$v^{2} - u^{2} = 2$ ×a×s

where, v = final velocity = 25 m/s

u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/ $s^{2}$

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

a= acceleration = 0.375 m/ $s^{2}$

Putting values we have F=375 N

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Eduardwww [97]

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3 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give

Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0

3 years ago

a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr

Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

$f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}}$ where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

$f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz$

5 0

3 years ago

8TH GRADE SCIENCE I NEED NOW DO NOT SKIP

yan [13]

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

=50(7.8)= 390N

7 0

3 years ago