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love history [14]
4 years ago
9

Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

. then . . . because. . . .” Be sure to answer the lesson question, “How do the factors of temperature and surface area affect the rate of chemical reactions?”
Physics
2 answers:
Paha777 [63]4 years ago
6 0

Answer:

If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

grandymaker [24]4 years ago
3 0
Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur. 
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PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?
Naily [24]

Answer: the frequency is every 27.322 days

6 0
3 years ago
Short-term memory is active, while long-term memory is:
Afina-wow [57]

Answer:

the answer is b. reflective

6 0
3 years ago
To withstand ‘g-forces’ of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a ‘human ce
Marina86 [1]

Answer:

34.292 m/s

Explanation:

When plane dives the gravitational acceleration becomes 10 times extra i-e

acceleration a = g= 10g's=10 ×9.8= 98 m/sec2

radius of dive circle = 12 m

v= speed= ?

Using

a= \frac{v^{2} }{r}

==> v^{2} = a × r = 98 × 12 = 1176 (\frac{m}{s}) ^{2}

==> v= \sqrt{1176(\frac{m}{s} )^{2} }

==> v= 34.292 m/s

4 0
3 years ago
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe
tia_tia [17]

Answer:

(1) 14.12 m/s

Explanation:

Given:

  • u = initial speed of the ball = 16 m/s
  • \theta = angle of the initial speed with the horizontal axis = 28^\circ
  • y_i = initial height of the ball from where Julie throws the ball = 1.5 m
  • y_f = final position of the ball where Sarah catches the ball = 1.5 m

Let us assume the following:

  • u_x = horizontal component of the initial speed
  • u_y = vertical component of the initial speed
  • a_x = horizontal acceleration of the ball
  • a_y = vertical acceleration of the ball

The given problem is projectile motion. When the ball is thrown from the air with a speed of 16 m/s at an angle 28 degree with the horizontal axis. When the ball is in the air, it experiences an only gravitational force in the downward direction if we ignore air resistance on the ball.

This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.

Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.

Let us find out the initial velocity horizontal component of the velocity of the ball. which is given by:

u_x = u\cos 28^\circ = 16\times \cos 28^\circ = 14.12\ m/s

As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.

Hence, the horizontal component of the ball's velocity when the ball is caught by Sarah is 14.12 m/s.

6 0
3 years ago
Read 2 more answers
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the thro
koban [17]

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

           

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

4 0
3 years ago
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