All categories

3 years ago

Grade 11 question paper or memo for 2015

Physics

2 answers:

denis-greek [22]3 years ago

5 0

What your saying doesn't make sense.

stira [4]3 years ago

4 0

Answer:

wym?

sry i dont comprehend

Explanation:

You might be interested in

If the bar is initially vertical, determine the angle of tilt of the bar when the 20 kipkip load is applied. Express your answer

Levart [38]

Answer:

Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the

6 0

4 years ago

What happens to water when it changes to ice?

mamaluj [8]

When water changes to ice, its density decreases and its volume increases.

7 0

3 years ago

Clothes dryer uses about 3:00 amps of current from a 240-volt line how much power does it use

kirill115 [55]

Electrical power is defined as

P = I * V

I = 3 amperes

V = 240 volts

Power = 3 * 240

Power = 720 Watts. Answer

4 0

4 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

Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe

laila [671]

m₁ = mass of sample of copper = m₂ = mass of sample of aluminum = 5 g

T = initial temperature of copper = initial temperature of aluminum

T₁ = final temperature of copper

T₂ = final temperature of aluminum

c₁ = specific heat of copper = 0.09 cal/g°C

c₂ = specific heat of aluminum = 0.22 cal/g°C

Since both receive same amount of heat, hence

Q₁ = Q₂

m₁ c₁ (T₁ - T) = m₂ c₂ (T₂ - T)

(5) (0.09) (T₁ - T) = (5) (0.22) (T₂ - T)

T₁ - T = (2.44) (T₂ - T)

Change in temperature of copper = (2.44) change in temperature of aluminum

hence the correct choice is

c. The copper will get hotter than the aluminum.

4 0

3 years ago

Read 2 more answers