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2 years ago

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Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant condit

ioning B. determinism C. self-efficacy D. self-worth

2 answers:

Darya [45]2 years ago

4 0

Answer hopefully = C. self-efficacy

Rzqust [24]2 years ago

4 0

Answer:

C

Explanation:

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A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

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2 years ago

Amy swims 500 m (0.5 km) in 6 minutes (0.1 hour ) . What was her speed in kilometers per hour ?

Scorpion4ik [409]

60 minutes = 1h
500/x = 10/100
She swam 5 kilometers per hour.

5 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

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2 years ago

Compare the signs of ƒ for lenses and mirrors.

STALIN [3.7K]

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

3 0

2 years ago

Read 2 more answers

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass $m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}$

$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

3 years ago