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3 years ago

Two friends are pushing a stalled,

Physics

1 answer:

Yuri [45]3 years ago

5 0

Answer:

91.03 N

Explanation:

consider the way the forces are acting on the car , in the graph

assume the car is accelerating to the right

we can apply Newton's second law of motion (F = ma )

→ F = ma

Fpush + 107 = 1610 * 0.123

Fpush = 91.03 N

second guy is applying 91.03 N force on the car

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The x-coordinates of two objects moving along the x-axis are given as a function of time t. x1 = (4 m/s)t and x2 = −(159 m) + (2

SpyIntel [72]

The x-coordinate of the first object is
x₁(t) = (4 m/s)t

The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²

The distance between the two objects is
x(t) = x₂ - x₁
= - 159 + 24t - t² - 4t
= -t² + 20t - 159

Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
= -[(t - 10)² - 100] - 159
= -(t-10)² - 59

This parabola has a vertex at (-10, -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.

The two graphs confirm that the analysis is correct.

Answer: The closest approach is 59 m.

3 0

3 years ago

A 53 kg person is being dragged in their sleeping bag to the lake by a 401 N force at an angle of 30°

masha68 [24]

In the horizontal direction, the forces acting on the person are

• friction with magnitude f, opposing motion, and

• the horizontal component of the pulling force (itself with mag. p ) with mag. p cos(30º), in the direction of motion.

There is no friction in the vertical direction, so we omit any discussion of the vertical forces.

By Newton's second law, we then have

p cos(30º) - f = m a cos(30º)

where m is the person's mass, and a is their acceleration so that a cos(30º) is the magnitude of the horizontal component of acceleration. The person is pulled by a force of p = 401 N, so solve for f :

(401 N) cos(30º) - f = (53 kg) (0.59 m/s²) cos(30º)

f ≈ 320 N

5 0

3 years ago

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test

emmasim [6.3K]

Answer:

(a) Frequency of the standing wave is 614.28 Hz.

(b) Frequency of the standing waves is 1228.56 Hz.

Explanation:

Frequency of standing wave in the case of one end open and other end closed of pipe (also known as stopped pipe) is given by the relation :

$f_{n} =n\frac{v}{4L}$ .....(1)

Here $f_{n}$ is the frequency of nth harmonic, v is speed of wave, L is length of the pipe and n is the harmonic which only takes values 1,3,5.. and so on.

(a) Given :

Speed of sound in air, v = 344 m/s

Length of tube, L = 14 cm = 0.14 m

n = 1

Substitute these values in equation (1).

$f_{1} =1\times\frac{344}{4\times0.14}$

f₁ = 614.28 Hz

(b) In this case, half of the tube is filled with water. So,

Length of tube having air, L = (0.14)/2 = 0.07 m

Substitute the suitable values in equation (1).

$f_{1} =1\times\frac{344}{4\times0.07}$

f₁ = 1228.56 Hz

7 0

3 years ago

What’s the acceleration if the average velocity is 3.5 and the time is 8.7

Monica [59]

Vf = 0 + 3.5•8.7
= 30.45 m/s

6 0

3 years ago

Energy transferred as heat always moves from an object

nata0808 [166]

Well, if we are being technical, yes and the only reason I say yes is because the sun is a hot gas rock. And it gives off UV rays which is heat on earth. So, yes. It does

3 0

3 years ago