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2 years ago

10

An unspecified force causes a 0.20-kg object to accelerate at 0.40 m/s2. If 0.30 kg is added to the 0.20-kg object and the force

remains the same, what is the acceleration (in m/s2) of the heavier object? (NEVER include units with the answer to a numerical question.)

1 answer:

Naddik [55]2 years ago

4 0

Answer:

a = 0.16

Explanation:

given,

mass of the object 1 = 0.2 kg

mass of the object 2 = 0.3 kg

acceleration when force is on 0.2 kg = 0.4 m/s²

acceleration when both mass are combine = ?

F = m a

F = 0.2 × 0.4

F = 0.08 N

force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg

F = m a

$a = \dfrac{F}{m}$

$a = \dfrac{0.08}{0.5}$

a = 0.16 m/s²

the acceleration acting when both the body is attached is a = 0.16

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Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof

bagirrra123 [75]

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

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1 year ago

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

1 year ago

a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0

2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

4 0

1 year ago