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Lubov Fominskaja [6]
3 years ago
8

If you increase the force on an item without changing the mass you increase acceleration? true or false?

Physics
1 answer:
rusak2 [61]3 years ago
5 0
(F)(M)=A
Force times Mass equals Acceleration.
The answer is TRUE. 
If the mass increases the number on the left side of the equation increases, thus increasing the right side as well.
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How much force does it take to accelerate a 50.8 kg person at 3.50 m/s^2?
Vikki [24]

Apply Newton's second law to the person's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

m = 50.8kg, a = 3.50m/s²

Plug in and solve for F:

F = 50.8(3.50)

F = 178N

8 0
2 years ago
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
Jo wants to find out about floating and sinking. She puts a rubber duck and a bar of soap in a
maks197457 [2]

Answer:

the soap sinks because it is more dense than the duck.

7 0
3 years ago
A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

Therefore, the spring constant is equal to 320 N/m

3 0
3 years ago
As speed (velocity) increases, potential energy increases true or false
Lyrx [107]

Answer:

true

Explanation:

3 0
2 years ago
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