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4 years ago

Select all that apply.

Physics

2 answers:

murzikaleks [220]4 years ago

8 0

D multiply force

Have a wonderful day !

Deffense [45]4 years ago

8 0

Answer:

Multiply force

Explanation:

Simple machines are used to multiply machines. There are several types of simple machines. For example : lever, pulley, wedge, screw etc.

A machine is used to increase the output force by doing small amount of output. The ratio of output force to the input force is called the mechanical advantage of a machine.

Hence, the correct option is (4) " multiply force ".

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Gamma rays are high-frequency EM waves and radio waves are low-frequency. Which characteristics are the same for both waves?

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C, they are transverse and travel at the speed of light

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4 years ago

The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·

soldier1979 [14.2K]

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

$Q_1 = m C_i \Delta T$

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

$\Delta T=0 C-(-20 C)=20^{\circ}C$ is the change in temperature of the ice

Substituting,

$Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal$

Now we have to find the amount of heat needed to melt the ice, which is

$Q_2 = m \lambda_f$

where

m = 150 g is the mass of the ice

$\lambda_f = 80 cal/g$ is the latent heat of fusion

Substituting,

$Q_2 = (150 g)(80 cal/g)=12,000 cal$

So the total heat required is

$Q_3 = 1500 cal + 12,000 cal = 13,500 cal$

2. 3750 cal

The additional amount of heat required to heat the water to 25°C is

$Q_4 = m C_w \Delta T$

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

$\Delta T=25 C-0 C=25^{\circ}C$ is the change in temperature

Substituting,

$Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal$

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

$Q_1 = m C_i \Delta T$

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

$\Delta T=0 C-(-20 C)=20^{\circ}C$ is the change in temperature of the ice

Substituting,

$Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal$

Now we have to find the amount of heat needed to melt the ice:

$Q_2 = m \lambda_f$

where

m = 80 g is the mass of the ice

$\lambda_f = 80 cal/g$ is the latent heat of fusion

Substituting,

$Q_2 = (80 g)(80 cal/g)=6,400 cal$

Finally, the amount of heat required to heat the water to 25°C is

$Q_3 = m C_w \Delta T$

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

$\Delta T=25 C-0 C=25^{\circ}C$ is the change in temperature

Substituting,

$Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal$

So the total heat required is

$Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal$

4. No

Explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.

4 0

3 years ago

If a simple machine reduces the strength of a force, what must be increased?

Katyanochek1 [597]

Your answer should be:
<span>a. the distance that the force moves</span>

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3 years ago

For a block to move down an inclined plane what force has to be the greatest?

Hatshy [7]

Answer:

D) True. This is what creates the body weight

Explanation:

Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)

X axis

Wx -fr = ma

Y Axis

N - Wy = 0

With trigonometry we can find the components of weight

sin θ = Wₓ / W

cos θ = $W_{y}$ / W

Wₓ = W sin θ

$W_{y}$ = W cos θ

W sin θ - fr = ma

From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body

Let's examine the answers

A False This force does not apply because it is not a spring

B) False. It is balanced at all times with the component (Wy) of the weight

C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower

D) True. This is what creates the body weight

E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points

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4 years ago

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

3 years ago