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stepan [7]
3 years ago
12

A student conducts an experiment to test how the temperature affects the amount of sugar that can dissolve in water. In the expe

riment, she uses 100 milliliters of water in each trial and stirs for five minutes each time.
What is the independent variable in this experiment?

A. the amount of water

B. the temperature of the water

C. the amount of sugar
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0
B. The temperature of the water.

The independent variable is the variable that is changed to affect the dependent variable. In this instance, the temperature of the water is being changed to affect the amount of sugar that dissolves.
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In situations involving equal masses, chemical reactions produce less energy than what reactions?
Liula [17]

in situations involving equal masses, chemical reactions produce less energy than nuclear reactions.

7 0
3 years ago
Read 2 more answers
A mom pushes her 19.3 kg daughter on the swing. If she gives her an initial velocity of 7.5 m/s at the bottom of the swing and t
kiruha [24]

Answer:

3.17333333333? I hope I get it right

Explanation:

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3 0
3 years ago
To practice Problem-Solving Strategy 8.1 for circular-motion problems. A cyclist competes in a one-lap race around a flat, circu
Anna007 [38]

Answer:

Explanation:

distance travelled

s = 2πR

= 2 X 3.14 X 140

= 880 m

final velocity = v

initial velocity = u

distance travelled = s

time = 60 s

s = ut + 1/2 at²

880 = .5 x a x 60²

a = .244 m/s²

final velocity v = at

= .244 x 60

= 14.66

centripetal acceleration at final moment

v² /R

14.66 X 14.66 / 140

= 1.53 m/s⁻²

1.53 m/s²

this is centripetal acceleration which acts towards the centre.

tangential acceleration calculated a _t = .244

redial acceleration ( centripetal ) = 1.53

Resultant acceleration

R²= 1.53² + .244 ²

R = 1.55 m/s²

total force = 1.55 x 76

= 118 N  

5 0
3 years ago
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
Xelga [282]

Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

6 0
2 years ago
3 Alpha Centauri is approximately 4.3 light years
sergey [27]

Answer:

40.68 trillion kilometers.

Explanation:

Multiply 4.3 × 9.46, the product which you get is your answer in trillion km

4 0
3 years ago
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