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**The graph above shows the force on an object** of mass M **as a function of time**. For the **time** interval 0 to 4 s, the total **change** in the **momentum** of the **object** is. answer choices 40 kgm/s ..

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4. The **graph** **above** **shows** **the** **force** **on** **an** **object** **of** mass M as a **function** **of** **time**. For **the** **time** interval 0 to 4 s, the total change in the momentum of the **object** is (**A**) 40 kg m/s (B) 20 kg m/s (C) 0 kg m/s (D) —20 kg m/s (E) indeterminable unless the mass M of the **object** is known 5. A 2 kg **object** initially moving with a constant velocity is.

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The** graph above shows** the force acting on an object as a function of time. The** change in momentum of the object** from me to lis a. 2 Ft b. Ft c.0.5 Ft d. 0.25 Ft e. 0 3. The force Fexeded on a ball during a collision with a wall is given as a function of time by the equation F (t) = ct- BE? where a = 100 N/s and 3 = 4000 N'S The ball first. **force**, we need to get hold of a 1 kg mass, have **the force act** on it somehow, and then measure the acceleration of the mass. The magnitude of the acceleration tells us the magnitude of **the force**; the direction of motion of the mass tells us the direction of **the force**. Fortunately, there are easier ways to measure forces. In addition to causing acceleration, forces cause **objects** to.

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**The graph** below **shows** the number of pages of the book left to read, y, after x weeks: A **graph** titled Lisas Book Reading **shows** Number of Weeks on the x-axis and Number of Pages Left on the Physics A **force** F1 of magnitude 6.90 units acts **on an object** at the origin in a direction θ = 54.0° **above** the positive x-axis.

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The graph above shows the force on an object of mass M as a function of time. For the time interval 0 to 4 s, the total change in the momentum of the object is answer** choices 40 kgm/s 20 kgm/s 0 kgm/s -20 kgm/s indeterminable unless the**.

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**The** **graph** **shows** **the** x-directed **force** F, **acting** **on** **an** **object** **as** **a** **function** **of** **the** position x of the **object**. For each numbered interval given, find the work W; done on the **object**. 1) from x = Om tox = 2.10m x (m) 2) from x = 3.00 m to x = 6.10m Wr= J 3) from x = 7.00 m to x = 9.30m.

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**The graph shows the force on an object** of mass M **as a function of time**. For the **time** interval 0 to 4 s, the total **change** in the **momentum** of the **object** is (A) 40 kg m/s (B) 20 kg m/s (C) 0 kg m/s (D) – 20 kg m/s (E) indeterminable unless the mass M of the **object** is known. This is known as the impulse-**momentum change** theorem.

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A particle of mass m is subjected to a **force** F. **change** in **force** with **time** is given by **the graph** in the question. **change in momentum** of the particle is given by the area under the curve. **Change in momentum** is equal to $\begin{align} & =\dfrac{1}{2}\times 2\times 6+\left( -3 \right)\times 2+3\times 4 \\ & =6-6+12 \\ & =12 \\ \end{align}$ So, the.

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**The****graph****above****shows****the****force****acting****on****an****object****as****a****function****of****time**.**The**change in momentum of the**object**from**time**0 to t is (**A**) 2Ft (B) Ft (C) F 2 (D) FI 4 (E) zero Question:**Force**Ques**Time**2 22. The**graph****above****shows****the****force****acting****on****an****object****as****a****function****of****time**.**The graph shows the force on an object**of mass M**as a function of time**. For the**time**interval 0 to 4 s, the total**change**in the**momentum**of the**object**is (A) 40 kg m/s (B) 20 kg m/s (C) 0 kg m/s (D) – 20 kg m/s (E) indeterminable unless the mass M of the**object**is known. This is known as the impulse-**momentum change**theorem.- Since the area under the line is really just multiplying
**force**and**time**, the area must be equal to the impulse**acting**on the**object**. Example 1: For**the graph**shown**above**, assume that it**shows**a constant**force**of 25 N**acting**over a 10 s period**of time**. Determine the impulse. Since area under the line is equal to impulse... Area=lw Area=25 10 **The graph shows the force on an object**of mass M**as a function of time**. For the**time**interval 0 to 4 s, the total**change in momentum**of the**object**is answer choices.**The graph above shows the force acting on an object as a function of time. The change in momentum**of the**object**from**time**0 totis a. 2 Ft b. Ft c.0.5 Ft d. Question: 3 kg 5 kg 2 kg ...- The impulse equals the area under a
**force**versus**time graph**, and this is extremely useful to know, because now in this section, where**the force**was varying, we can still use this, we can just find the impulse by determining the area under that curve. And by area under the curve, we mean from the line curve, in general, to the x-axis, which, in ...