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dangina [55]
3 years ago
15

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?​

Physics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

F = 8 N

Explanation:

The question says "The body is subjected to a force with a moment of 0.4 N × m shoulder - 5 cm. What is the magnitude of this force?

Given that,

Moment/Torque, \tau=0.4\ N-m

Distance moved, d = 5 cm = 0.05 m

We need to find the magnitude of this force. We know that, the torque acting on an object is given by :

\tau=Fd\\\\F=\dfrac{\tau}{d}\\\\F=\dfrac{0.4\ N-m}{0.05\ m}\\\\F=8\ N

So, the magnitude of force is equal to 8 N.

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When are all the forces acting upon an object balanced?
Rudik [331]

Answer:

When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.

Explanation:

7 0
3 years ago
When are magnets attracted and when are they repelled.
vova2212 [387]
Magnets are attracted when each of the different sides, most commonly known as "North" and "South", are facing each other. They repel when North and North, or South and South are facing each other.
3 0
4 years ago
Can you tell from the coefficient of restitution whether a collision has added kinetic energy to a system, taken some away, or l
Maurinko [17]
For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.

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6 0
3 years ago
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the
Over [174]

Answer:

Part a)

a= 0.32 m/s^2

Part b)

F_c = 3.6 N

Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

F_f = 3.45 N

Now we know by Newton's II law

F_{net} = ma

so we have

F_p - F_f = (m_a + m_b) a

9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

a= 0.32 m/s^2

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

3 0
3 years ago
A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i
dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

h=ut+\frac{at^2}{2}

here Initial vertical velocity is 0

8.3=\frac{gt^2}{2}

t^2=1.69

t=1.301 s

Horizontal distance

R=u\times t

R=30\times 1.301=39.04 m

7 0
3 years ago
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