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3 years ago

Use the data from Table D of your Student Guide to answer the questions. Be sure to round your answers to the

Physics

2 answers:

Vikentia [17]3 years ago

8 0

Answer:

2kg cart = 16.0

4kg cart =8.1

6kg cart =5.3

Explanation:

Romashka [77]3 years ago

7 0

Answer:

Hi, you haven't provided the Table D of your Study Guide which contains the other variables needed to calculate the acceleration.

Therefore, I will just explain how to calculate the acceleration when we have the mass and the net force on the object.

Explanation:

For this example, we can use the Newton's second law of motion states that an object acceleration (a) is a function of the mass (m) of the object and the net force (F) on the object. Thus:

F = ma; a = F/m

(a) When the mass of the cart = 2 kg; if the net force is 32 kgcm/s^2

Then, a = F/m = 32/2 = 16.0 cm/s^2

(b) When the mass of the cart = 4 kg

Then, a = F/m = 32/4 = 8.0 cm/s^2

Then, a = F/m = 32/6 = 5.3 cm/s^2

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3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s

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What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

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4 years ago

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

3 years ago