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marusya05 [52]
3 years ago
5

Hii! help asap. i’ll give brainliest thanks!

Physics
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

a i believe

Explanation:

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MATHPHYS Physics help
Nesterboy [21]

Answer:

360 J/kg/°C

Explanation:

Heat lost by metal = heat gained by water

-m₁ C₁ (T − T₁) = m₂ C₂ (T − T₂)

-(0.059 kg) C (27.4°C − 221°C) = (0.409 kg) (4186 J/kg/°C) (27.4°C − 25°C)

(11.4 kg°C) C = 4109 J

C = 360 J/kg/°C

8 0
4 years ago
Read 2 more answers
He sees a mouse sniffing along at 0.2 m/s but it hears and starts to scurry for safety. In just 4.7 s it speeds up to 1.6 m/s. F
NISA [10]

Answer:

0.30 m/s²

Explanation:

Given:

v₀ = 0.2 m/s

v = 1.6 m/s

t = 4.7 s

Find: a

a = (v − v₀) / t

a = (1.6 m/s − 0.2 m/s) / 4.7 s

a = 0.30 m/s²

8 0
4 years ago
What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed
Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above;  the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on  a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0
3 years ago
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
An object is dropped from a height of 50 meters. Ignoring air resistance, how long doe take to hit the ground?
Arte-miy333 [17]

here given that object is dropped from height h = 50 m

So here we can say

initial speed is ZERO

acceleration is due to gravity

now in order to find time to reach the ground we can use kinematics

y = v_i*t + \frac{1}{2}at^2

now plug in all values in it

50 = 0 + \frac{1}{2} \times 9.8(t^2)

t = 3.2 s

so it will take 3.2 s to reach the ground

3 0
3 years ago
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