Heat _________________ throughout our environment ________________ of the _______________

A 2 kg toy cart and a 6 kg toy cart have a spring compressed between them. When the spring expands, it sends the 2 kg toy cart o

harkovskaia [24]

Answer:

The speed of second toy cart is 4 m/s.

(c) is correct option

Explanation:

Given that,

Mass of first toy cart = 2 kg

Mass of second toy cart = 6 kg

Speed of first toy cart = 12 m/s

We need to calculate the speed of second toy cart

Using formula of momentum

$m_{1}v_{1}=m_{2}v_{2}$

Where, m₁ = mass of first toy cart

m₂ = mass of second toy cart

v₁ = velocity of first toy cart

v₂ = velocity of second toy cart

Put the value into th formula

$2\times12=6\times v_{2}$

$v_{2}=\dfrac{2\times12}{6}$

$v_{2}=4\ m/s$

Hence, The speed of second toy cart is 4 m/s.

(c) is correct option

4 0

4 years ago

SONAR stands for "sound navigation and ranging,” and it is used to map and explore the ocean floor.

lara31 [8.8K]

SONAR stands for "sound navigation and ranging,” and it is used to map and explore the ocean floor.

3 0

3 years ago

A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall buildin

Bas_tet [7]

Answer:

Explanation:

Consider downward displacement first .

downward displacement h = 38 m .

downward acceleration = g

downward initial velocity u = 0

h = ut + 1/2 g t ²

38 = 0 + 9.8 t ²

t = 1.97 s

During this period , there will be horizontal displacement with initial velocity u = 10 m /s and acceleration a = 1.6 m /s²

s = ut + 1/2 a t²

= 10 x 1.97 + .5 x 1.6 x 1.97²

= 19.7 + 3.10

= 22.8 m

3 0

3 years ago

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie

telo118 [61]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The uncertainty in inverse frequency is $\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

Explanation:

From the question we are told that

The value of the proportionality constant is $k = 5 \frac{Hz }{T}$

The strength of the magnetic field is $B = 20 \ T$

The change in this strength of magnetic field is $\Delta B = 3 \ T$

The magnetic field is given as

$B = \frac{k}{\frac{1}{w} }$

Where $w$ is frequency

The uncertainty or error of the field is given as

$\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ]$

The uncertainty in inverse frequency is given as

$\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}$

$\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }$

substituting values

$\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }$

$\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

6 0

3 years ago

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago

Heat _________________ throughout our environment ________________ of the ________________.

Heat _ throughout our environment of the .