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hoa [83]
3 years ago
13

Consider the electric force between a pair of charged particles a certain distance apart. By Coulomb's Law: If instead, the char

ge of both particles is doubled, the force is _____.
a. unchanged
b. halved
c. doubled
d. quadrupled
Physics
1 answer:
Westkost [7]3 years ago
5 0

Answer:

the answer is d

Explanation:

from F=Q1Q2

r².

then wen both charges are doubled

F=2Q1 ×2Q2

r²

F=4Q1Q2

r²

then 4 has to be the factor multiplied by the other part of the equation for it to balance

4F=4Q1Q2

r²

so theoretically the 4 on the LHS can cancel the 4 on the RHS

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A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m
Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

   Horizontal displacement = 40.5 m

   Time  

         t=\frac{40.5}{28.64}=1.41s          

   Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

    Time to reach the goal posts 40.5 m away = 1.41 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

    Height of throw = 1.4 m

    Height traveled by ball = 2.83 m

    Total height = 2.83 + 1.4 = 4.23 m

    When the ball goes to first base it will be 4.23 m high.

8 0
3 years ago
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0
3 years ago
Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
Natalija [7]

Let say for every 5 s of time interval the speed will remain constant

so it is given as

v(mi/h)   16    21    23    26    33    30     28

now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s

so here we will have

v(ft/s)      23.5    30.8     33.73     38.13     48.4     44     41.1

now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

so here it will cover 1298.1 ft distance in 30 s interval of time

4 0
3 years ago
Read 2 more answers
For phyics again , pls !!!
CaHeK987 [17]

Answer: same

Explanation: They both weigh a kilogram and there is no friction

3 0
2 years ago
What is the reaction force if a girl pulls on a cow?
Papessa [141]
Answer : B) The cow pulls back on the girl.

From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
8 0
3 years ago
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