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Stells [14]
2 years ago
12

Sarah rides her horse with a constant speed of 12km/h. How far can she travel in 3 hours?

Physics
1 answer:
Law Incorporation [45]2 years ago
3 0
  • Speed=12km/h
  • Time=3h

\\ \sf\Rrightarrow Distance=Speed\times Time

\\ \sf\Rrightarrow Distance=12(3)

\\ \sf\Rrightarrow Distance=36km

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
Which type of energy is the original source for the energy that food molecules can provide?
Svet_ta [14]

Answer:

Chemical potential energy

Explanation:

which is used to for any form of energy.

4 0
3 years ago
Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?
Stolb23 [73]

Answer:

200 cal/^{\circ}C

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

Q=C\Delta T

where

Q is the heat supplied

C is the heat capacity of the object

\Delta T is the change in temperature

In this problem we have:

Q=10,000 cal is the energy supplied

\Delta T=75C-25C=50C is the change in temperature of the object

Therefore, the heat capacity of the object is:

C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C

6 0
3 years ago
P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
3 years ago
A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume
lara [203]

Answer:

what is the upwards force?

5 0
2 years ago
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