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romanna [79]
3 years ago
8

A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them

using a spring scale. The table shows the different masses used in the experiment and the force required to pull each mass. The student concluded that more force was required to pull heavier objects. What comment would you make regarding his conclusion? A) No clear relation can be observed between mass and force from the data. B) There is a direct proportion between the mass and force listed in the table. C) Gravity should have been taken into account while performing the experiment. D) There is an inverse proportion between the mass and force listed in the table.
Mass (kg) Force (N)

5 25

10 50

15 75

20 100
Physics
2 answers:
Gwar [14]3 years ago
5 0

Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

saul85 [17]3 years ago
4 0

Answer: Correct answer is B.

Hope this helps.

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A motorist, traveling east on an open highway, sets his cruise control at 90.0 km/h. How far will he travel in 0.75 hours?
dimaraw [331]

Answer:

<h2>The answer is 67.5 km</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 90 × 0.75

We have the final answer as

<h3>67.5 km</h3>

Hope this helps you

3 0
2 years ago
Read 2 more answers
Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist
andreyandreev [35.5K]

Answer:

a) It is moving at 43.15\frac{m}{s^{2}} when reaches the ground.

b) It is moving at 101.44\frac{m}{s^{2}} when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

W=K_f-K_i (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

K=\frac{mv^2}{2} (2)

with m the mass and v the velocity.

Using (2) on (1):

W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

W=Fd=wd=mgd (4)

Using (4) on (3):

mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}

v_f=43.15\frac{m}{s^{2}}

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

-mgd=-\frac{mv_i^2}{2}

Solving for initial velocity (when the boulder left the volcano):

v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}

v_i=101.44 \frac{m}{s^{2}}

3 0
3 years ago
Starting velocity: 50 m/s
Taya2010 [7]

Please find attached photograph for your answer. Please do comment whether it is useful or not

6 0
3 years ago
What step does the Rin PRICES stand for? Why is this step important?
posledela

Explanation:

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6 0
3 years ago
The Heaviside function H is defined by H(t)={0 if t&lt;0, 1 if t≥0 It is used in the study of electric circuits to represent the
Studentka2010 [4]

Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

so we see that the Heaviside function "switches on" whent=0, and remains switched on when t>0

If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

f(t) = H(t-5)

The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

V(t)= 240V* H(t-5)

I have made a sketch for you, and added it as attachment.  

5 0
3 years ago
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