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3 years ago

A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

Physics

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)$

so we have

$H = 1.47 m$

Part b)

Similarly when spring is compressed by 4 cm

then we have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)$

so we have

$H = 5.94 m$

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A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?

vovangra [49]

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

5 0

3 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

<em>Substitute </em> $F_{r}$ <em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

4 0

2 years ago

1. How many seconds in 1 year?

Nezavi [6.7K]

Answer:

3.154e+7

Explanation:

8 0

3 years ago

Read 2 more answers

Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

7 0

3 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago