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alexandr1967 [171]
3 years ago
7

An astronaut is doing repairs on the international Space Station. She throws a hammer into space with a velocity of 3m/s. How mu

ch force is needed to keep the 5kg hammer moving at 3m/s through space?
Physics
2 answers:
Rainbow [258]3 years ago
3 0

Answer:

The force is 15N

Explanation:

The formula is Force= mass × velocity.

From the question mass is 5kg, velocity is 3m/s.

F= 5×3

F= 15Newton.

Therefore the force is 15N.

anastassius [24]3 years ago
3 0

Answer:

7.5 N

Explanation:

Given data :

velocity ( V ) = 3m/s

mass of hammer ( m ) = 5 kg

Determine the force needed to keep hammer moving through space

The force needed to keep the hammer moving in space can be calculated

F = ma  ----- (1 )

m = 5 kg

a = velocity / time

velocity = displacement / time  = 3 m/s  hence displacement after 1 second = 3m

now we have to determine the acceleration using the relationship  below

V^2 =  V1^2 + 2(a)(d)

3^2 = 0 + 2(a) ( 3 )

 9 = 6a   ∴  a = 9/6 = 1.5 m/s^2

back to equation 1

F = ma

  = 5 * 1.5 =  7.5 N

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot
KiRa [710]

Answer:

T_{f} = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

      Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

     Q₁ = 2.00 10⁻³ 2.26 10⁶

     Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

     Q coffee = M c_{e} ( T_{f} -T_{i})

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

    Qc = - Q₁

    M ce (T_{f} -T_{i}) = - Q₁

The coffee dough left in the cup after evaporation is

    M = 250 -2 = 248 g = 0.248 kg

   T_{f} -Ti = -Q1 / M c_{e}

   T_{f} = Ti - Q1 / M c_{e}

Since coffee is essentially water, let's use the specific heat of water,

    c_{e}= 4186 J / kg ºC

Let's calculate

     T_{f} = 90.0 - 4.52 103 / (0.248 4.186 103)

     T_{f} = 90- 4.35

     T_{f} = 85.65 ° C

     T_{f} = 85.7 ° C

5 0
3 years ago
Your friend asks you for a glass of water and you bring her 5 milliliters of water. Is this more or less than what she was proba
Eddi Din [679]
Probably not what you were expecting... the average bottle of water is 24 ounces. 5 milliliters is about the amount of water in a spoon. Hope this helps!!!
3 0
3 years ago
Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force
kkurt [141]
Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=
7 0
3 years ago
Read 2 more answers
An automatic clothing drier spins at 51.6 rev/min. If the radius of the drier drum is 30.5cm, how fast (in m/s) is the drier dru
Jlenok [28]

Answer:

1.648 m/s

Explanation:

1 revolution equals 2pi radians.

Calculate the angular velocity by taking 2pi x v, then divide by 60 seconds.

To convert this to m/s, simply take this answer and multiply it by 0.305m (a.k.a. the radius of the circle).

3 0
3 years ago
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