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Alona [7]
3 years ago
5

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

Physics
1 answer:
ioda3 years ago
3 0
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
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Darya [45]

Gap junctions in the intercalated discs allow impulses to be spread across the heart more quickly. This is because gap junctions allow particles/signals to pass through, thus making cells with gap junctions more able to interact.

One more thing—you posted this in the physics section rather than biology.

8 0
3 years ago
The speed of an aircraft is sometimes expressed as a mach number: mach 1 means that the speed is equal to the speed of sound. if
mars1129 [50]
<span>The speed of sound needs to be given, in the proper form. This will allow for the proper conversion (namely, a multiplication by the Mach rate) to find the actual speed that the aircraft is traveling, compared to how fast sound travels.</span>
8 0
4 years ago
Một vôn kế đang đo hiệu điện thế ở thang đo có giá trị mỗi độ chia là 0,1 V. Nếu kim chỉ thị nằm giữa vạch 5,5 V và 5,6 V thì đọ
kakasveta [241]

Answer:

5.6V nha bạn

Explanation:

5 0
3 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
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