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2 years ago

When measuring current, always place any type of meter in

2 answers:

5 0

In series with the circuit, so for it pass the current to be mensured.

Letter A

If you notice any mistake in my english, please let me know, because i am not native.

Delicious77 [7]2 years ago

4 0

Answer:

Series with the circuit

Explanation:

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With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s

Dima020 [189]

initial velocity=u=20km/h=5.5m/s
Acceleration=a=8m/s^2
Time=t=10s

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2$

$\\ \sf\longmapsto s=55+4(5.5)^2$

$\\ \sf\longmapsto s=55+4(30.25)$

$\\ \sf\longmapsto s=55+121$

$\\ \sf\longmapsto s=176m$

4 0

2 years ago

Calculate the Force of Gravity acting on an object that has a mass of 1.3 kg. The object is on Earth so use 9.8m/s2 for the acce

sweet-ann [11.9K]

To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.

Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.

The solution is C. 12.7 N.

7 0

2 years ago

Which of the following is NOT one of the four elements of emotion?

Nezavi [6.7K]

Answer:

Attention

Explanation:

Hello there, fellow peer! The answer to question is attention. Let's say someone is the control. The behavioral expression is an element of expression, so the control will feel emotions. Subjective Experience is when someone felt the way you feel and they are trying to help you. That is a type of emotion which can lead to empathy for you. This is also not the answer. Physiological Arousal is also not the answer because this is when you can feel what someone else is feeling and you try to give them therapy.

Using the process of elimination, our answer is therefore attention.

4 0

3 years ago

a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an

11111nata11111 [884]

When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school

6 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago