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eduard
3 years ago
5

Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the

second force is applied at 30° to the plane of the door. Which force exerts the greater torque about the door hinge?
Physics
1 answer:
Leviafan [203]3 years ago
3 0

Answer:

in first case the torque is maximum.

Explanation:

Torque is defined as the product of force and the perpendicular distance.

τ = F x d x Sinθ

In case A: the angle between force vector and the distance vector is 90 so torque is

τ = F x d

In case B: the angle between force vector and the distance is 30°.

τ = F x d x Sin30

τ = 0.5 Fd

So the torque is maximum in first case.

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a fan is rotating clockwise and its acceleration has a positive sign. is the angular velocity of the fan speeding up, slowing do
balu736 [363]

Answer:

The angular velocity is slowing down.

Explanation:

  • By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
  • If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.
5 0
2 years ago
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil
laila [671]

Answer:

W_n_e_t=7.648512 \approx 7.6J

K.E=0.8J

v=0.7844645406 \approx 0.78m/s

Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

Distance traveled 48cm=>0.48m

Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

F=0.28*2.6*9.8

F=7.1344N

Generally work done on the pitcher is mathematically given as

W_n_e_t=W_f+W_F

W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

W_n_e_t=4.224-3.424512

W_n_e_t=0.799488\approx 0.8J

b)Generally K.E can be given mathematically as

K.E= W_n_e_t

Therefore

K.E=0.8J

c)Generally the equation for kinetic energy is mathematically represented by

K.E=1/2mv^2

0.8=1/2mv^2

Velocity as subject

v=\sqrt{\frac{K.E*2}{m} }

v=\sqrt{\frac{0.8*2}{2.6} }

v=0.7844645406 \approx 0.78m/s

6 0
3 years ago
What morbid structure traditionally has thirteen steps?
tatiyna

Answer:

A gallow

Explanation:

6 0
3 years ago
What is the current that reverses direction in a regular patter called
Oxana [17]

Answer:

Current that reverses direction in the regular pattern is called an alternating current, abbreviated as 'AC'.

Explanation:

hope this helps!

4 0
3 years ago
If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
creativ13 [48]

Answer:

-112.876J

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

16KNO_3(s) + 24C(s) + S_8(s)    \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)

Let us define P - V work as;

w_{pv} = - P_{external}  \triangle Volume

where  \triangle (Volume) = (V_{final} - V_{initial})

External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2

K_2S is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  CO_2  and  N_2 into grams using their respective molar masses.

0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

7 0
3 years ago
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