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3 years ago

Look at the graph. If the price of peaches increases, what can be expected?

Physics

2 answers:

Fittoniya [83]3 years ago

5 0

There will be less peaches soled

max2010maxim [7]3 years ago

5 0

Answer:

The correct answer is B) Fewer peaches will be sold.

Explanation:

Hope this helps! :)

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Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.

Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0

2 years ago

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

2 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

2 years ago

A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0

3 years ago

En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)

Virty [35]

Answer:

I only speak English

Explanation:

I'm sorry can you type it in English

7 0

3 years ago