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Brut [27]
3 years ago
11

A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What typ

e of acceleration Is this?​
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

Acceleration = 4.8 m/s²

Explanation:

Given:

Change in velocity = 19 m/s

Change in time = 4 s

Find:

Acceleration

Computation:

Acceleration = Change in velocity / Change in time

Acceleration = 19/4

Acceleration = 4.8 m/s²

Positive acceleration

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A dog travels 18 meters south across the backyard in 11 seconds. What is the dog's speed?
o-na [289]
The dog’s speed is
A) 0.61 m/s
4 0
3 years ago
A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
3 years ago
If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre
Dmitrij [34]
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m

Now, just substitute these into the formula:
</span>F =  \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }<span>

</span>
3 0
3 years ago
Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi
77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0
3 years ago
Suppose a blood vessel's radius is decreased to 87% of its original value by plaque deposits and the body compensates by increas
dexar [7]

Answer:

The pressure difference will increase by the factor of 1.75

Explanation:

For constant flow rate, coefficient of viscosity, length of the vessel and the pressure difference is inversely proportional to the fourth power of the radius of the blood vessel

Apply the principle of Poiseuille’s law.

Q = (P2 - P1)/R

Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.

3 0
3 years ago
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