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DENIUS [597]
3 years ago
7

1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?

Physics
1 answer:
shepuryov [24]3 years ago
7 0

1.

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N


2.

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg


3.

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N


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A wire along the z axis carries a current of 6.4 A in the z direction Find the magnitude and direction of the force exerted on a
almond37 [142]

Answer:

The magnetic force will be 0.256 N in +y direction.                                          

Explanation:

It is given that, a wire along the z axis carries a current of 6.4 A in the z direction. Length of the wire is 8 cm. It is placed in uniform magnetic field with magnitude 0.50 T in the x direction.

The magnetic force in terms of length of wire is given by :

F=I(L\times B)\\\\F=ILB\\\\F=6.4\times 0.08\times 0.5\\\\F=0.256\ N

For direction,

F=I k(L\times Bi)\\\\F=Fj

So, the magnetic force will be 0.256 N in +y direction.

7 0
3 years ago
If Earth was with no tilt, would we still have seasons at all? If so, how would they be different?
murzikaleks [220]

Answer:

If earth had no tilt, we would have no seasons.

Explanation:

As stated in the answer, if the earth had no tilt we wouldn't have seasons. The earth all around the globe would maintain the same temperature,

And due to the no tilt it would also change our orbit to a bit larger slant, in January when we are at our closest to the sun we WOULD have a mini summer. For the North and South Pole, they would remain cold.

8 0
3 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre
Yuliya22 [10]

Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

<u>Step 4:</u>  What work is done by the truck on the gravel?  

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

5 0
3 years ago
What is (9x10^9)(2.6x10^-6)(1.4x10^-6) / 36
Natali [406]

Answer:

0.00091

Explanation:

(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36

(9,000,000,000) (0.0000026) (0.0000014) /36

|

23,400(0.0000014) /36

|

0.03276 /36

|

0.00091

3 0
3 years ago
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