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Zepler [3.9K]
3 years ago
9

PLEASE HELP! I'LL GIVE BRAINLEST​

Physics
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

You might be interested in
A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0
3 years ago
this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85
mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

3 0
1 year ago
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10
marissa [1.9K]

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

7 0
3 years ago
Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
ICE Princess25 [194]
The answer is A-40N.
8 0
3 years ago
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

4 0
3 years ago
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