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andrezito [222]
3 years ago
12

Which describes how scientific models are used? (2 points) They allow scientists to make predictions. They prevent scientists fr

om performing experiments. They make it easier for scientists to avoid using data. They prevent people from criticizing the work of scientists.
Physics
1 answer:
luda_lava [24]3 years ago
4 0
I think the correct answer would be the first option. Scientific models are used to allow scientists to extrapolate or interpolate data. Or in other words, allow them to predict values without performing more experiments. However, they are obtained by experiments first. 
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A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)
Alekssandra [29.7K]

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

Magnetic field = 0.150 T

Speed v= 4.40\times10^{6}\ m/s

We need to calculate the electric field strength

Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

4 0
3 years ago
A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebou
Vsevolod [243]

Answer:

\Delta p=2.27\frac{kg\cdot m}{s}

Explanation:

The momentum change is defined as:

\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)

Taking the downward motion as negative and the upward motion as positive, we have:

v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)

Replacing (2) and (3) in (1):

\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}

5 0
3 years ago
A sprinter has a mass of 80 kg and a KE of 4000 J. What is the sprinter’s speed?
djyliett [7]
There you go.

Hope this helps!

8 0
3 years ago
Read 2 more answers
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
3 years ago
A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘
Alenkasestr [34]

Answer: a. 198.6J b. - 198.6J

Explanation: Parameters given:

m = 15kg

g = 9.8m/s²

∅ = 12°

a. Work done by the force Fp on the cart if the ramp is 6.5m long.

Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N

Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J

b. The work done by the force mg on the cart.

Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance

= 15kg x -9.8m/s² x Sin12° x 6.5m

= - 198.6J

4 0
3 years ago
Read 2 more answers
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