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svp [43]
1 year ago
8

in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse

Physics
1 answer:
Greeley [361]1 year ago
8 0

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

Learn more about impulse here

brainly.com/question/16980676

#SPJ4

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Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen??
Reptile [31]

Answer: The formula of Newtons second law of motion is F=MA so therefore it would be written like this Force = Mass X Acceleration

F = 5 x 2

F = 10 N

6 0
3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
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Variations in the reactive properties of different organic molecules are most closely associated with _____.
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Answer:

the presence or absence of functional groups

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The functional group is the group of atoms that characterize a chemical function and that have well-defined characteristic properties.

In organic chemistry, the functional group is a set of submolecular structures, characterized by a specific elementary connectivity and composition that confers specific chemical reactivity to the molecule that contains them. These structures replace the hydrogen atoms lost by saturated hydrocarbon chains. Aliphatic, or open chain, groups are usually represented generically by R (alkyl radicals), while aromatic ones, or derivatives of benzene, are represented by Ar (aryl radicals).

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