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3 years ago

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From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h

ave after 4s?
A. 50m / s
B. 60m / s
C. 70m / s
D. 80m / s

2 answers:

stepan [7]3 years ago

4 0

The answer:

A) 50 m/s

Yuki888 [10]3 years ago

3 0

Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

Find: v

v = at + v₀

v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

v² = (30 m/s)² + (40 m/s)²

v = 50 m/s

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A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0

3 years ago

Help please science kids

IceJOKER [234]

It helps because it's being transported blah blah whatever the last person said when you first asked this question

6 0

3 years ago

a train engineer started the train from a standstill and sped up to 5 meters per second, she then rounded a corner at a constant

olasank [31]

for acceleration we can define that rate of change in velocity is know as acceleration

So whenever velocity of train is changing with time we can say train is accelerating

Now here if initially train is standstill then after some time its speed is 5 m/s

so here the train is accelerated first time

Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time

After this train chugged around a curve with same speed 10 m/s

SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration

SO this is accelerated Third time

Then its speed decreases and it comes to speed of 5 m/s from 10 m/s

So here it is acceleration of train for Fourth time

Then finally train comes to stop so again its speed changed from 5 m/s to 0

so this is acceleration of train Fifth time

So total train will accelerate 5 times in whole path

3 0

3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

4 years ago

13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V a

Lisa [10]

Answer:

The output power is 2 kW

Explanation:

It is given that,

Number of turns in primary coil, $N_p=250$

Number of turns in secondary coil, $N_s=500$

Voltage of primary coil, $V_p=200\ V$

Current drawn from secondary coil, $I_s=5\ A$

We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}$

$\dfrac{250}{500}=\dfrac{200}{V_s}$

$V_s=400\ V$

So, the power output is :

$P_s=V_s\times I_s$

$P_s=400\ V\times 5\ A$

$P_s=2000\ watts$

or

$P_s=2\ kW$

So, the output power is 2 kW. Hence, this is the required solution.

8 0

3 years ago