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Andreas93 [3]
3 years ago
12

A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground​

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

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A bicyclist travels at the speed of 25 kilometers per hour for 5 hours. How far does the bicyclist go?
mixas84 [53]

Answer:

125 kilometers

Explanation:

Speed= 25km/h

Time = 5 h

Distance= Speed x Time

= 25 x 5

= 125 km

6 0
2 years ago
Read 2 more answers
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
5 What is the kinetic energy of a 28 kg car moving at a velocity of 7 m/s O 1963 98 686 J O 350 nocter diagram, at which point w
Shtirlitz [24]

Answer:

98 joules

Explanation:

1/2 of 28 = 14

14 multiply by 7= 98 joules

4 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
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